UC-NRLF 


SB    270    fiOl 


OF 


An  Electrical  Library. 

By  PROF.  T.  O'CONOR  SLOANE. 

How  to  become  a  Successful  Electrician. 

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Electricity  Simplified. 

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Electric  Toy  Making,  Dynamo  Building,  etc. 

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Arithmetic  of  Electricity. 

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Standard  Electrical  Dictionary. 

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NORMAN  W.  HENLEY  &  CO.,  Publishers, 

132  Na»»au  Street,  New  York. 


Arithmetic  of  .Electricity 


A  PRACTICAL  TREATISE  ON  ELECTRICAL  CALCULATIONS  OF 
ALL  KINDS  REDUCED  TO  A  SERIES  OF  RUIZES,  ALI,  OF  THE 
SIMPLEST  FORMS,  AND  INVOLVING  ONLY  ORDINARY  ARITH- 
METIC, EACH  RULB  ILLUSTRATED  BY  ONE  OR  MORE  PRAC- 
TICAL PROBLEMS,  WITH  DETAILED  SOLUTION  OF  EACH, 
FOLLOWED  BY  AN  EXTENSIVE  SERIES  OF  TABLES. 

BY 

T.   O'CONOR   SLOANE,   A.M.,   E.M.,  Ph.D. 

»s 

AUTHOR   OF 

Standard  Electrical  Dictionary,  Electricity  Simplified, 
Electric  Toy  Making,  etc. 


TWENTIETH  EDITION 


NEW  YORK : 

THE  NORMAN  W.  HENLEY  PUBLISHING  COMPANY 

132  NASSAU  STREET 

1909 


COPYRIGHTED,  1891, 

BY 
NORMAN  W.  HENLEY  &  CO. 


COPYRIGHTED,   1903, 

BY 
NORMAN  W.  HENLEY  &  CO. 


COPYRIGHTED,   1909, 

BY 

THE  NORMAN  W.  HENLEY  PUBLISHING  CO. 
\ 


PREFACE. 


The  solution  of  a  problem  by  arithmetic,  although 
in  some  cases  more  laborious  than  the  algebraic  method, 
gives  the  better  comprehension  of  the  subject.  Arith- 
metic is  analysis  and  bears  the  same  relation  to  algebra 
that  plane  geometry  does  to  analytical  geometry.  Its 
power  is  comparatively  limited,  but  it  is  exceedingly 
instructive  in  its  treatment  of  questions  to  which  it 
applies. 

In  the  following  work  the  problems  of  electrical  en- 
gineering and  practical  operations  are  investigated  on 
an  arithmetical  basis.  It  is  believed  that  such  treatment 
gives  the  work  actual  value  in  the  analytical  sense,  as 
it  necessitates  an  explanation  of  each  problem,  while 
the  adaptability  of  arithmetic  to  readers  who  do  not 
care  to  use  algebra  will  make  this  volume  more  widely 
available. 

In  electricity  there  is  much  debatable  ground,  which 
has  been  as  far  as  possible  avoided.  Some  points  seem 
quite  outside  of  the  scope  of  this  book,  such  as  the  intro- 
duction of  the  time-constant  in  battery  calculations. 
Again  the  variation  in  constants  as  determined  by  dif- 
ferent authorities  made  a  selection  embarrassing.  It  is 
believed  that  some  success  has  been  attained  in  over- 
coming or  compromising  difficulties  such  as  those  sug- 
gested. 


383491 


iv  PREFACE. 

Enough  tables  have  ^een  introduced  to  fill  the  limits 
of  the  subject  as  here  treated. 

The  full  development  of  electrical  laws  involves  the 
higher  mathematics.  One  who  would  keep  up  with  the 
progress  of  the  day  in  theory  has  a  severe  course  of 
study  before  him.  In  practical  work  it  is  believed  that 
such  a  volume  as  the  Arithmetic  of  Electricity  will 
always  have  a  place.  We  hope  that  it  will  be  favorably 
received  by  our  readers  and  that  their  indulgence  will 
give  it  a  more  extended  field  of  usefulness  than  it  can 
pretend  to  deserve. 


PREFACE  TO  TWENTIETH  EDITION. 

The  steady  progress  of  electrical  science  in  conjunc- 
tion with  a  continued  demand  for  this  work  have  made 
advisable  a  revision  and  extension  of  this  book. 

The  author  feels  that  in  the  matter  which  has  been 
added  much  more  could  have  been  said  on  the  subjects 
treated  of,  but,  since  a  full  exposition  of  each  theme 
would  alone  fill  a  volume,  it  is  hoped  that  the  practical 
value  of  the  rules,  etc.,  will  atone  for  the  brevity  of  the 
text. 

In  the  preparation  of  this  edition  the  author  would 
express  his  indebtedness  to  A.  A.  Atkinson's  excellent 
work  on  Electrical  and  Magnetic  Calculations  and  also 
to  the  instruction  papers  of  the  Electrical  Engineering 
Course  of  the  International  Correspondence  School  of 
S'cranton,  Pa.  He  would  also  express  his  thanks  to 
Henry  V.  A.  Parsell,  for  his  valued  advice  and  assistance 
in  the  preparation  of  the  manuscript. 

THE  AUTHOB. 


CONTENTS. 


CHAPTER  L 


INTRODUCTORY. 


Space,  Time,  Force,  Resistance,  Work,  Energy,  Mass 
and  Weight. — The  Fundamental  Units  of  Dimension, 
and  Derived  Units,  Geometrical,  Mechanical,  and  Elec- 
trical.— C.  G.  S.  and  Practical  Electrical  Units. — Nomen- 
clature.— Examples  from  Actual  Practice 9 


CHAPTER   IL 
OHM'S  LAW. 


General  Statement.— Six  Rules  Derived  by  Transpos- 
ition from  the  Law.— Single  Conductor  Closed  Circuits. 
—Batteries  in  Opposition.— Portions  of  Circuits.— Di- 
vided Circuits,  with  Calculation  of  Currents  Passed  by 
Each  Branch,  and  of  their  Combined  Resistance 13 


CHAPTER  III. 

RESISTANCE  AND    CONDUCTANCE. 

Resistance  of  Different  Conductors  of  the  same  Mater- 
ial.— Relations  of  Wires  of  Equal  Resistance. — Ratio  of 
Resistance  of  Two  Conductors.— Specific  Resistance.— 
Universal  Rule  for  Resistances.— Resistance  of  Wires 
Referred  to  Weight.— Conductance.— Ohm's  Law  ex- 
pressed in  Conductance .26 


vi  CONTENTS. 

CHAPTER   IV. 

4  POTENTIAL    DIFFERENCE. 

Drop  of  Potential  in  Leads,  and  Size  of  Same  for  Mul- 
tiple Arc  Connections. — Diminishing  Size  of  Leads 
Progressively 38 

CHAPTER  V. 

CIRCULAR  MILS. 

The  Mil.— The  Circular  Mil  as  a  Unit  of  Area.— Cir- 
cular Mil  Rules  for  Resistance  and  Size  of  Leads 43 


CHAPTER  VI. 

SPECIAL  SYSTEMS. 

Three  Wire  System. — Rules  for  Calculating  Leads  in 
Same. — Alternating  Current  System. — Ratio  of  Conver- 
sion.— Size  of  Primary  Wire. — Converter  Winding 46 


CHAPTER  VII. 

WORK    AND    ENERGY. 

Energy  and  Heating  Effect  of  the  Current. — Differ- 
ent Rules  Based  on  Joule's  Law. — The  Joule  or  Gram- 
Calorie.— Quantity  of  Heat  Developed  in  an  Active  Cir- 
cuit in  a  Unit  of  Time.  — Watts  and  Amperes  in  Rela- 
tion to  Time. — Specific  Heat. — Heating  of  Wire  by  a 
Current.— Safety  Fuses.— Work  of  a  Current.— Elec- 
trical Horse-Power.— Duty  and  Efficiency  of  Electrical 
Generators .  50 


CONTENTS.  Mil 

CHAPTER  VIII. 

BATTEBIES. 

Arrangement  of  Battery  Cells. — General  Calcula- 
tions of  Current. — Rules  for  Arrangement  of  Cells 
in  a  Battery. — Battery  Calculations  for  Specified 
Electromotive  Force  and  Current. — Efficiency  of 
Batteries. — Chemistry  of  Batteries. — Calculation 
of  Voltage. — Work  of  Batteries. — Efficiency  of  Bat- 
teries, to  Calculate. — Chemicals  Consumed  in  a 
Battery. — Decomposition  of  Compounds  by  a  Bat- 
tery.— Electroplating  66 


CHAPTER  IX. 

ELECTRO-MAGNETS,    DYNAMOS    AND    MOTORS. 

The  Magnetic  Field  and  Lines  of  Force. — Per- 
meance and  Reluctance. — Magnetizing  Force  and 
the  Magnetic  Circuit. — General  Rules  for  Electro- 
Magnets  and  Ampere-Turns  for  Given  Magnetic 
Flux. — Magnetic  Circuit  Calculations. — Leakage  of 
Lines  of  Force. — Example  of  Calculation  of  a  Mag- 
netic Circuit. — Dynamo  Armatures. — Voltage  and 
Capacity  of  Armatures. — Drum  Type  Closed  Cir- 
cuit Armatures. — Field  Magnets  of  Dynamos. — The 
Kapp  Line  82 


CHAPTER  X. 

ELECTRIC  RAILWAYS. 

Sizes  of  Feeders. — Power  to  Move  Cars 109 


viii  CONTENTS. 

CHAPTER  XI. 

ALTERNATING    CURRENTS. 

Self-Induction    115 

CHAPTER  XII. 

CONDENSERS. 

Page  121 

CHAPTER  XIII. 

DEMONSTRATION    OF   RULES. 

Some  of  the  Principal  Rules  in  the  Work  Demon- 
strated    127 

CHAPTER  XIV. 

NOTATION    IN    POWERS    OF    TEN. 

The  Four  Fundamental  Operations  of  Addition, 
Subtraction,  Multiplication,  and  Division  in  Powers 
of  Ten  136 

TABLES. 

A  Collection  of  Tables  Needed  for  the  Operations 
and  Problems  Given  in  the  Work  .  .  139 


ARITHMETIC  OF  ELECTRICITY. 


CHAPTER  L 

INTRODUCTORY. 

SPACE  is  the  lineal  distance  from  one  point  to 
another. 

Time  is  the  measure  of  duration. 

Force  is  any  cause  of  change  of  motion  of  matter. 
It  is  expressed  practically  by  grams,  volts,  pounds 
or  other  unit. 

Resistance  is  a  counter-force  or  whatever  opposes 
the  action  of  a  force. 

Work  is  force  exercised  in  traversing  a  space 
against  a  resistance  or  counter-force.  Force  multi- 
plied by  space  denotes  work  as  foot-pounds. 

Energy  is  the  capacity  for  doing  work  and  is 
measurable  by  the  work  units. 

Mass  is  quantity  of  matter. 

Weight  is  the  force  apparent  when  gravity  acts 
upon  mass.  When  the  latter  is  prevented  from 
moving  under  the  stress  of  gravity  its  weight  can 
be  appreciated. 


10  ARITHMETIC  OF  ELECTRICITY. 

Physical  and  Mechanical  calculation,  are  based  on 
three  fundamental  units  of  dimension,  as  follows: 
the  unit  of  time — the  second,  T;  the  unit  of  length 
— the  centimeter,  L;  the  unit  of  mass — the  gram, 
M.  Concerning  the  latter  it  is  to  be  distinguished 
from  weight.  The  gram  is  equal  to  one  cubic  centi- 
meter of  water  under  standard  conditions  and  is 
invariable;  the  weight  of  a  gram  varies  slightly  with 
the  latitude  and  with  other  conditions. 
•  Upon  these  three  fundamental  units  are  based  the 
derived  units,  geometrical,  mechanical  and  electrical. 
The  derived  units  are  named  from  the  initials  of 
their  units  of  dimension,  the  C.  G.  S.  units,  indi- 
cating centimeter-gram-second  units. 

In  practical  electric  calculations  we  deal  with 
certain  quantities  selected  as  of  'convenient  size 
and  as  bearing  an  easily  defined  relation  to  the 
fundamental  units.  They  are  called  practical 
units. 

The  cause  of  a  manifestation  of  energy  is  force; 
if  of  electromotive  energy,  that  is  to  say  of  electric 
energy  in  the  current  form,  it  is  called  electromotive 
force,  E.  M.  F.  or  simply  E.  or  difference  of  poten- 
tial D.  P.  What  this  condition  of  excitation  may 
be  is  a  profound  mystery,  like  gravitation  and  much 
else  in  the  physical  world.  The  practical  unit  of 
E.  M.  F.  is  the  VOLT,  equal  to  one  hundred  mil- 
lions (100,000,000)  0.  G.  S.  units  of  E.  M.  F.  The 
last  numeral  is  expressed  more  briefly  as  the  eighth 


INTRODUCTORY.  11 

power  of  10  or  108.  Thus  the  volt  is  defined  as 
equal  to  108  C.  G.  S.  units  of  E.  M.  F. 

This  notation  in  powers  of  10  is  used  throughout 
C.  G.  S.  calculations.  Division  by  a  power  of  10  is 
expressed  by  using  a  negative  exponent,  thus  10~* 
means  ioooagoog»  Tlie  exj)onent  indicates  the  number 
of  ciphers  to  le  placed  after  1. 

When  electromotive  force  does  work  a  current  is 
produced.  The  practical  unit  of  current  is  the 
AMPERE,  equal  to  &  C.  G.  S.  unit,  or  10'1  C.  G.  S. 
unit,  •&  being  expressed  by  10"1. 

A  current  of  one  ampere  passing  for  one  second 
gives  a  quantity  of  electricity.  It  is  called  the 
COULOMB  and  is  equal  to  10'1  C.  G.  S.  units. 

A  coulomb  of  electricty  if  stored  in  a  recipient 
tends  to  escape  with  a  definite  E.  M.  F.  If  the 
recipient  is  of  such  character  that  this  definite  E. 
M.  F.  is  one  volt,  it  has  a  capacity  of  one  FARAD 
equal  to  TOOOO^OOOO  or  10~9  C.  G.  S.  unit. 

A  current  of  electricity  passes  through  some 
substances  more  easily  than  through  others.  The 
relative  ease  of  passage  is  termed  conductance.  In 
calculations  its  reciprocal,  which  is  resistance,  is 
almost  universally  used.  A  current  of  one  ampere 
is  maintained  by  one  volt  through  a  resistance  of 
one  practical  unit.  This  unit  is  called  the  OHM  and 
is  equal  to  109  C.  Gr.  S.  units. 

Sometimes,  where  larger  units  are  wanted,  the  pre- 
fix deka,  ten  times,  lieka*  one  hundred  times,  Tcilo,  one 


12  ARITHMETIC  OF  ELECTRICITY. 

thousand  times,  or  mega,  one  million  times  are  used, 
as  dekalitre,  ten  liters,  kilowatt,  one  thousand  watts, 
megohm,  one  million  ohms. 

Sometimes,  where  smaller  units  are  wanted,  the 
prefixes,  deci,  one  tenth,  centi,  one  hundredth,  milli, 
one  thousandth,  micro,  one  millionth,  are  used.  A 
microfarad  is  one  millionth  of  a  farad. 

For  the  concrete  conception  of  the  principal  units 
the  following  data  are  submitted. 

A  DanielFs  battery  maintains  an  E.  M.  F.  of  1.07 
volt.  A  current  which  in  each  second  deposits 
.00033  grains  copper  (by  electro-plating)  is  of  one 
ampere  intensity  and  from  what  has  been  said  the 
copper  deposited  by  that  current  in  one  second  cor- 
responds to  one  coulomb.  A  column  of  mercury  one 
millimeter  square  and  106.24  centimeters  long  has 
a  resistance  of  one  ohm  at  0°  0.  The  capacity  of  the 
earth  is  itUooo  farad.  A  Leyden  jar  with  a  total 
coated  surface  of  one  square  meter  and  glass  one 
mm.  thick  has  a  capacity  of  -h  microfarad.  The 
last  is  the  more  generally  used  unit  of  capacity. 

These  practical  units  are  derived  from  the  0.  G-. 
S.  units  by  substituting  for  the  centimeter  (C.)  one 
thousand  million  (109)  centimeters  and  for  the  gram, 
the  one  hundred  thousand  millionth  (10~n)  part  of  a 
gram. 


CHAPTER  II. 
OHM'S  LAW. 

THIS  law  expresses  the  relation  in  an  active 
electric  circuit  (circuit  through  which  a  current  of 
electricity  is  forced)  of  current,  electromotive  force, 
and  resistance.  These  three  factors  are  always  pres- 
ent in  such  a  circuit.  Its  general  statement  is  as 
follows  : 

In  an  active  electric  circuit  the  current  is  equal  to 
the  electromotive  force  divided  by  the  resistance. 

This  law  can  be  expressed  in  various  ways  as  it  is 
transposed.  It  may  be  given  as  a  group  of  rules,  to 
be  referred  to  under  the  general  title  of  OHM'S  LAW. 

Rule    1.     The  current  is  equal  to  the  electromotive 
force  divided  by  the  resistance.  C  =  — 

R 

Rule  2.   The  electromotive  force  ts  equal  to  the  cur- 
rent multiplied  by  the  resistance.  E  =  C  R 

Rule  3.   The  resistance  is  equal  to  the  electromotive 
force  divided  by  the  current. 

~  C 

Rule  4.  The  current  varies  directly  with  the  electro* 
motive  force  and  inversely  with  the  resistance. 


14  ARITHMETIC  OF  ELECTRICITY. 

Rule  5.  The  resistance  varies  directly  \vitli  tlic  elec* 
troiiiotivc  force  and  inversely  with  the  current. 

Rule  6.  The  electromotive  force  varies  directly  with 
the  current  and  with  the  resistance. 

This  law  is  the  fundamental  principle  in  most 
electric  calculations.  If  thoroughly  understood  it 
will  apply  in  some  shape  to  almost  all  engineering 
problems.  The  forms  1,  2,  and  3  are  applicable  to 
integral  or  single  conductor  circuits;  when  two  or 
more  circuits  are  to  be  compared  the  4th,  5th  and  6th 
are  useful.  The  law  will  be  illustrated  by  examples. 

SINGLE  CONDUCTOR  CLOSED  CIKCUITS. 

These  are  circuits  embracing  a  continuous  con', 
ducting  path  with  a  source  of  electromotive  force 
included  in  it  and  hence  with  a  current  continually 
circulating  through  them. 

EXAMPLES. 

A  battery  of  resistance  3  ohms  and  E.  M.  F.  1.07 
volts  sends  a  current  through  a  line  of  wire  of  55 
ohms  resistance  ;  what  is  the  current? 

Solution  :  The  resistance  is  3  +  55  =  58  ohms. 
By  rule  1  we  have  for  the  current  J^/-  giving  .01845 
Ampere. 

Note. — A  point  to  be  noticed  here  is  that  whatever  is 
included  in  a  circuit  forms  a  portion  of  it  and  its  resistance 
must  be  included  therein.  Hence  the  resistance  of  the 
battery  has  to  be  taken  into  account.  The  resistance  of  a 
battery  or  generator  is  sometimes  called  internal  resistance 


OHM'S  LAW.  15 

to  distinguish  it  from  the  resistance  of  the  outer  circuit, 
called  external  resistance.  Kesistance  in  general  is 
denoted  by  R,  electromotive  force  by  E,  and  current  by 
C. 

A  battery  of  R  2  ohms;  sends  a  current  of  .035 
ampere  through  a  wire  of  R  48  ohms;  what  is  the 
E.  M.  F.  of  the  battery? 

Solution:  The  resistance  is  48  +  2  =  50  ohms.  By 
Rule  2  we  have  as  the  E.  M.  F.  50  X  .035  =  1.75 
volts. 

A  maximum  difference  of  potential  E.  M.  F.  of 
30  volts  is  maintained  in  a  circuit  and  a  current  of 
191  amperes  is  the  result;  what  is  the  resistance  of 
the  circuit? 

Solution:  By  Rule  3  the  resistance  is  equal  to 
ohms. 


In  the  same  circuit  several  generators  or  gal- 
vanic couples  may  be  included,  some  opposing  the 
others,  i.  e.  connected  in  opposition.  All  such  can 
be  conceived  of  as  arranged  in  two  sets,  distrib- 
uted according  to  the  direction  of  current  produced 
by  the  constituent  elements,  in  other  words,  so  as 
to  put  together  all  the  generators  of  like  polarity. 
The  voltages  of  each  set  are  to  be  added  together  to 
get  the  total  E.  M.  F.  of  each  set. 

Rule  7.  Where  batteries  or  generators  are  In  opposi- 
tion, add  together  the  E.  M.  F  of  all  generators  of  like 
polarity,  thus  obtaining  two  opposed  E.  OT.  F.s.  Sub- 
tract the  smaller  E.  JJI.  F.  from  the  larger  E.  OT.  F.  to 


16 


ARITHMETIC  OF  ELECTRICITY. 


obtain  the  effective  i:.  Itt.  F.    Then  apply  Ohm's  la\v  on 
this  basis  of  E.  M.  F. 

It  will  be  understood  that  the  resistances  of  all 
batteries  or  generators  in  series  are  added  to  give 
the  internal  resistance. 

EXAMPLES. 

There  are  four  batteries  in  a  circuit:  Battery  No. 
1  of  2  volts,  y2  ohm;  Battery  No.  2  of  1.75  volts,  2 
ohms;  Battery  No.  3  of  1  volt,  1  ohm;  Battery  No. 
4  of  1  volt,  4  ohms  constants;  Batteries  1  and  4  are 
in  opposition  to  2  and  3.  What  are  the  eifective 
tattery  constants? 


Solution:  Voltage  =  (2  +  1)  —  (1.75  +  1)  =  .25  volt. 
Kesistance  =  ^  +  2+1  +  4  =  7^  ohms,  or  .25 
volt,  7^  ohms  constants. 

What  current  will  such  a  combination  produce  in 
a  circuit  of  5  ohms  resistance? 

Solution:  By  Ohm's  law,  Rule  1,  the  current  =* 
.25  •*•  (7^  +  5)  =  .02  amperes. 

A  battery  of  51  volts  E.  M.  F.  and  20  ohms  resist- 


S  LAW.  17 

ance  has  opposed  to  it  in  the  same  circuit  a  battery 
of  26  volts  E.  M.  F.  and  25  ohms  resistance.  A 
current  of  ^  ampere  is  maintained  in  the  circuit. 
What  is  the  resistance  of  the  wire  leads  and  con- 
nections ? 

Solution:  The  effective  E.  M.  P.  is  51  —  26  =  25 
volts.  By  Eule  3  we  have  25  •+•  i  =  200  ohms,  as  the 
total  resistance.  But  the  resistance  of  the  batteries 
(internal  resistance)  is  20  +  25  =  45  ohms.  The  re- 
sistance of  leads,  etc.  (external  resistance),  is  there- 
fore 200  —  45  =  155  ohms. 

PORTIONS  OF  CIRCUITS. 

All  portions  of  a  circuit  receive  the  same  current, 
but  the  E.  M.  F.,  in  this  case  termed  preferably  dif- 
ference of  potential,  or  drop  or  fall  of  potential, 
and  the  resistance  may  vary  to  any  extent  in  differ- 
ent sections  or  fractions  of  the  circuit.  Ohm's  Law 
applies  to  these  cases  also. 

EXAMPLES, 

An  electric  generator  of  unknown  resistance  main- 
tains a  difference  of  potential  of  10  volts  between  its 
terminals  connected  as  described.  The  terminals 
are  connected  to  and  the  circuit  is  closed  through  a 
series  of  three  coils,  one  of  100  ohms,  one  of  50 
ohms,  and  one  of  25  ohms  resistance.  The  connec- 
tions between  these  parts  are  of  negligibly  low  re- 


18  ABITHMETIC  OF  ELECTKICITY. 

sistance.  What  difference  of  potential  exists  be- 
tween  the  two  terminals  of  each  coil  respectively? 

Solution:  The  solution  is  most  clearly  reached  by 
a  statement  of  the  proportion  expressed  in  Rule  6, 
viz. :  The  electromotive  force  varies  directly  with  the 
resistance.  The  resistance  of  the  three  coils  is  175 
ohms;  calling  them  1,  2,  and  3,  and  their  differences 
of  potential  E1,  E2,  and  E8,  we  have  the  continued 
proportion,  175  :  100  :  50  :  25  ::  10  volts  :  E1 :  E2 :  E8. 
because  by  the  conditions  of  the  problem  the  total 
E.  M.  F.  =  10.  Solving  the  proportion  by  the  regu- 
lar rule,  we  find  that  E1  =  5.7,  E8  =  2.8  and  E3  =  1.4 
volts. 

The  same  external  circuit  is  connected  to  a 
battery  of  30  ohms  resistance.  The  difference  of 
potential  of  the  100  ohm  coil  is  found  to  be  30  volts. 
What  is  the  difference  of  potential  between  the  ter- 
minals of  the  battery,  and  what  is  the  E.  M.  F.  of 
the  battery  on  open  circuit,  known  as  its  voltage  or 
E.  M.  F.  (one  of  the  battery  constants)? 

Solution:  The  total  external  resistance  is  100  +  50 
+  25  =  175  ohms.  By  Rule  6,  we  have  100  :  175  ::  30 
volts  :  x  =  52^  volts,  difference  of  potential  between 
the  terminals  of  the  battery.  The  current  is  found 
by  dividing  (Rule  1),  the  difference  of  potential  of 
the  100  ohm  coil  by  its  resistance.  This  E.  M.  F.  is 
30.  The  current  therefore  is  •££$  amperes.  The  to- 
tal resistance  of  the  circuit  is  that  of  the  three  coils 
or  175  ohms  plus  that  of  the  battery  or  30  ohms,  a 


OHM'S  LAW.  19 

total  of  205  ohms.  To  maintain  a  current  of  ffo 
amperes  through  205  ohms  (Rule  2),  an  E.  M.  F.  is 
required  equal  to  i3&  X  205  volts  or  61^  volts. 

DIVIDED,  BRANCHED  OR  SHU.NT  CIRCUITS. 

A  single  conductor,  from  one  terminal  of  a  gener- 
ator may  be  divided  into  one  or  more  branches 
which  may  reunite  before  reaching  the  other  ter- 
minal. Such  branches  may  vary  widely  in  resist- 
ance. 

Rule  8.  In  divided  circuits,  eacli  branch  passes  a 
portion  of  a  current  inversely  proportional  to  its  re* 
sistance. 

EXAMPLES. 

A  portion  of  a  circuit  consists  of  two  conductors, 
A  and  B,  in  parallel  of  A  =  50,  and  B  =  75  ohms, 
respectively;  what  will  be  the  ratio  of  the  currents 
passing  through  the  circuit,  which  will  go  through 
each  conductor? 

Solution:  The  ratio  will  be  current  through  A  : 
current  through  B  ::  75  :  50,  which  may  be  ex- 
pressed fractionally,  &  :  T^. 

Where  more  than  two  resistances  are  in  parallel, 
the  fractional  method  is  most  easily  applied. 

Three  conductors  of  A  =  25,  B  =  50,  and  0  =  75 
ohms  are  in  parallel.  What  will  be  the  ratio  of  cur- 
rents passing  through  each  one? 

Solution:  Fractionally  AzBrC::^:^:^. 


20  ARITHMETIC  OF  ELECTRICITY. 

It  tile  9.  To  determine  the  amount  of  a  given  cur- 
rent that  will  pass  through  parallel  circuits  of  differ* 
ent  resistances,  proceed  as  follows :  Take  the  resist- 
ance of  each  branch  for  a  denominator  of  a  fraction 
having  1  for  its  numerator.  In  other  words,  for  each 
branch  write  down  the  reciprocal  of  its  resistance. 
Then  reduce  the  fractions  to  a  common  denominator, 
and  add  together  the  numerators.  Taking  this  sum 
of  the  numerators  for  a  new  common  denominator, 
and  the  original  single  numerators  as  numerators, 
the  new  fractions  will  express  the  proportional  cur- 
rents as  fractions  of  one.  If  the  total  amperage  is 
given,  it  is  to  be  multiplied  by  the  fractions  to  give 
the  amperes  passed  by  each  branch.  The  solution  can 
also  be  done  in  decimals. 


EXAMPLES. 

A  lead  of  wire  divides  into  three  branches;  No. 
1  has  a  resistance  of  10,000  ohms,  No.  2  of  39  ohms, 
and  No.  3  of  i  ohm.  They  unite  at  one  point. 
What  proportion  of  a  unitary  current  will  pass  each 
branch  ? 

Solution:  The  proportion  of  currents  passed  are 
as  rotary :  *V  :  %  or  3.  Reducing  to  a  common  de- 
nominator, these  become  T^Hhny  *  AWir :  V&°iW.  The 
proportions  of  the  numerators  is  the  one  sought  for; 
taking  the  sum  of  the  numerators  as  a  common  de- 
nominator, we  have  in  common  fractions  the  follow- 
ing proportions  of  any  current  passed  by  the  three 

branches.     No.   1,   n^mr;  No.    2,   Ttt***r;  No.    3, 
1170000 

118  OT5T9* 

Four  parallel  members  of  a  circuit  have  resistances 
respectively  of  25,  85,  90,  and  175  ohms;  express 


OHM'S  LAW.  21 

decimally  the  ratio  of  a  unitary  current  that  will 
pass  through  them. 

Solution:  The  ratio  is  as  &  :  £s  :  ir& :  rf?,  or  reduc- 
ing to  decimals  (best  by  logarithms),  .04  :  .011765  : 
.011111  :  .0057.  Adding  these  together,  we  have 
.068576,  which  must  be  multiplied  by  14.582  to  pro- 
duce unity.  Multiplying  each  decimal  by  14.58 
(best  by  logarithms),  we  get  the  unitary  ratio  as 
.5832  :  .17153  :  .1620  :  .08310,  whose  sum  is  1.0000. 

Unless  logarithms  are  used,  it  is  far  better  to  work 
by  vulgar  fractions. 

A  current  of  .71  amperes  passes  through  two 
branches  of  a  circuit.  One  is  a  lamp  with  its  con- 
nections of  115  ohms  resistance;  another  is  a  resist- 
ance coil  of  275  ohms  resistance.  What  current 
passes  through  each  branch? 

Solution:  The  proportions  of  the  current  are  as 
TIT  :  T&S  or  reduced  to  a  common  denominator  and  to 
their  lowest  terms  ^ftr  :  ^ff*.  Proceeding  as  before, 
and  taking  the  sum  of  the  numerators  (55  +  23  — 
78),  as  a  common  denominator,  we  find  that  the  lamp 
passes  H,  and  the  resistance  coil  H  of  the  whoie 
current.  Multiplying  the  whole  current,  .71  by  f$, 
we  get  nn  amperes,  or  i  ampere  for  the  lamp,  leaving 
.21  or  a  little  over  I  ampere  for  the  resistance  coil. 

Another  problem  in  connection  with  parallel 
branches  of  a  circuit  is  the  combined  resistance  of 
parallel  circuits.  This  is  not  a  case  of  summa- 


22  ARITHMETIC  OF  ELECTRICITY. 

tion,  for  it  is  evident  that  the  more  parallel  paths 
there  are  provided  for  the  current,  the  less  will  be 
the  resistance. 

Rule  1  0.  In  shunt  circuits,  the  resistance  of  the  com- 
bined shunts  is  expressed  by  the  reciprocal  of  the  sum 
of  the  reciprocals  of  the  resistances. 

EXAMPLE. 

Two  leads  of  a  50  volt  circuit  (leads  differing  in 
potential  by  50  volts),  are  connected  by  a  20  ohm 
motor.  A  50  ohm  lamp  and  1000  ohm  resistance 
coil  are  connected  in  parallel  or  shunt  circuit  there- 
with, what  is  the  combined  resistance?  and  the  total 
current? 

Solution:  The  reciprocal  of  resistance  is  conduc- 
tance, sometimes  expressed  as  mhos.  (Rule  19.) 
The  conductance  of  the  three  shunts  is  equal  to 
A  +  ifo  +  raW  mhos  =  riHhr  +  -Ms  +  irinr  =  yHv 
mhos.  The  reciprocal  of  conductance  is  resistance. 
The  combined  resistance  is  therefore  -W  ohms  = 
14.09  ohms.  The  current  is  Jj!L  or  3.5  amperes. 

Rule  11.  The  combined  resistance  of  two  parallel 
circuits  is  found  by  multiplying  the  resistance*  to- 
gether, and  dividing  the  product  by  the  sum  of  the  re- 
sistances. Where  there  are  several  circuits,  any  t\vo 
can  be  treated  thus,  and  the  result  combined  in  *he 
same  way  with  another  circuit,  and  so  on  to  getthe 


final  resistance. 


B  _ 
R  — 


^ 

r  x  rl 


EXAMPLE. 

Four  conductors  in  parallel  have  resistances  of 
100  —  50  —  27  —  19  ohms.  What  is  their  combined 
resistance? 


OHM'S  LAW.  23 

Solution:  Combining  the  first  and  second,  we 
have  1™^  =  33i  ohms.  Combining  this  with  the 
resistance  of  the  third  wire,  we  have  ^|^^  =  14.9 
ohms.  Combining  this  with  the  resistance  of  the 
fourth  wire,  we  have  ^$S  =  8.3  ohms.  The 
result  is,  of  course,  identical  by  whatever  rule 
obtained. 

Rale  12.  When  all  the  parallel  circuits  are  of  uni- 
form resistance,  as  in  multiple  arc  incandescent  light- 
ing, the  resistance  of  the  combined  circuits  is  found  by 
dividing  the  resistance  of  one  circuit  by  the  number  of 
circuits.  _  £ 

—  n 

EXAMPLES. 

There  are  fifty  lamps  of  100  ohms  resistance  each 
in  multiple  arc  connection.  What  is  their  com- 
bined resistance? 

Solution:  W  =  2  ohms. 

A  motor  can  take  3  amperes  of  currents  at 
30  volts  safely  without  burning  out  or  heating 
injuriously.  A  110  volt  incandescent  circuit  is  at 
hand.  The  motor  is  to  be  connected  across  the 
leads  so  as  to  receive  the  above  amperage.  A  shunt 
or  branch  of  some  resistance  is  carried  around  it, 
and  a  resistance  coil  intervenes  between  the  united 
branches  and  one  of  the  main  leads.  The  resistance 
of  the  coil  is  20  ohms.  What  should  the  resistance 
of  the  shunt  be? 


24  ARITHMETIC  OF  ELECTEICITY. 


A. 
-VWWW: 


UAA/WSAA) 

rfe.si.sta.njce 


Solution:  The  resistance  of  the  motor  (Ohm's 
Law,  Rule  3),  is  found  by  dividing  the  E.  M.  F.  by 
fthe  resistance  —  30  •*•  3  =  10  ohms.  By  Rule  5  the 
resistance  of  the  coil  in  series  (20  ohms)  must  be  to 
the  combined  (not  added)  resistance  of  the  motor 
and  shunt  coil,  as  110  —  CO  (total  voltage  minus 
voltage  for  motor)  :  30  (voltage  for  motor)  or  20  : 
x  ::  80  :  30  .  •.  x  =  7.5  combined  resistance  of  parallel 
or  shunt  coil  and  motor.  The  reciprocal  of  7.5 
(conductance,  Rule  19),  may  be  expressed  as  HSths  of 
the  combined  (in  this  case  added)  conductances  of 
shunt  coil  and  motor.  The  conductance  of  the 
motor  is  equal  to  the  reciprocal  of  10  which  may  be 
expressed  as  &  or  as  ^.  The  conductance  of  the 
shunt  coil  must  therefore  be  H#  ~~  ^  =  T¥O  =  ih> 
mho.  The  reciprocal  of  this  gives  the  resistance  of 
the  shunt  coil  which  is  30  ohms.  The  total  current 
going  through  the  system  by  Ohm's  law  is 


amperes.  The  resistance  of  the  shunt  coil  —  30  ohms 
—  is  to  that  of  the  motor  in  parallel  with  it  —  10 
ohms  —  as  the  current  received  by  the  motor  is  to 


OHM'S  LAW.  25 

that  received  by  the  coil,  a  ratio  of  30  :  10  or  3:1 
giving  3  amperes  for  the  motor  and  1  ampere  for 
the  coil.  This  is  a  proof  of  the  correctness  of 
operations. 

Two  conductors  through  which  a  current  is 
passing  are  in  parallel  circuit  with  each  other. 
One  has  a  resistance  of  600  ohms.  The  other  has  a 
resistance  of  3  ohms.  A  wire  is  carried  across  from 
an  intermediate  point  of  one  to  a  corresponding 
point  of  the  other.  It  is  attached  at  such  a  point  of 
the  first  wire  that  there  are  400  ohms  resistance  be- 
fore it  and  200  after  it.  Where  must  it  be  connected 
to  the  other  in  order  that  no  current  may  pass? 

Solution:  The  E.  M.  F.  up  to  the  point  of  con- 
nection of  the  bridge  or  cross  wire  is  to  the  total 
E.  M.  F.  in  the  600  ohm  wire  as  400:  600  or  as  2:  3. 
The  other  wire  which  by  the  conditions  has  the 
same  drop  of  potential  in  its  full  length  must  be 
divided  therefore  in  this  ratio.  The  bridge  wire 
must  therefore  connect  at  2  ohms  from  its  begin- 
ning, leaving  1  ohm  to  follow.  The  principle  here 
illustrated  can  be  proved  generally  and  is  the  Wheat- 
stone  Bi  jdge  principle. 


CHAPTER  III. 
KESISTANCE  AND  CONDUCTANCE. 

RESISTANCE  OF  DIFFERENT  CONDUCTORS  OF  THE 
SAME  MATERIAL. 

Conductors  are  generally  circular  in  section. 
Hence  they  vary  in  section  with  the  square  of  their 
diameters.  The  rule  for  the  resistance  of  conduc- 
tors is  as  follows: 

Rule  13.  The  resistance  of  conductors  of  identical 
material  varies  inversely  as  their  section,  or  if  of  circu- 
lar section  inversely  as  the  squares  of  their  diameters, 
and  directly  as  their  lengths. 

EXAMPLE. 

1.  A  wire  #,  is  30  mils  in  diameter  and  320  feet 
long;  another  b,  is  28  mils  in  diameter  and  315  feet 
long.  What  are  their  relative  resistances? 

Solution:  Calling  the  resistances  R*  :  Rb  we  would 
have  the  inverse  proportion  if  they  were  of  equal 
lengths  Rb  :  R*  ::  302  :  282  or  as  900  :  784.  Were 
they  of  equal  diameter  the  direct  proportion  would 
hold  for  their  lengths:  Rb  :  Ra  ::  315  :  320.  Com- 
bining the  two  by  multiplication  we  have  the  com- 
pound proportion  Rb  :  Ra  ::  900  X  315  :  784  X  320  or 
as  283,500  :  250,880,  or  as  28  :  25  nearly.  The 


RESISTANCE  AND  CONDUCTANCE.  27 

combined  proportions  could  have  been  originally 
expressed  as  a  compound  proportion  thus:  Kb  :  Ra  :: 
30*  X  315  :  282  X  320. 

For  wires  of  equal  resistance  the  following  is 
given. 

Rule  14.  The  length  of  one  wire  multiplied  by  the 
square  of  the  diameter  of  the  other  wire  must  equal  the 
square  of  its  own  diameter  multiplied  by  the  length  of 
the  other  If  their  resistances  are  equal.  Or  multiply  the 
length  of  the  first  wire  by  the  square  of  the  diameter  of 
the  second.  This  divided  by  the  length  of  the  second 
will  give  the  square  of  the  diameter  of  the  first  wire; 
or  divided  by  the  square  of  the  diameter  of  the  first  will 
give  the  length  of  the  second.  Id"  —  I'd8 

EXAMPLES. 

1.  There  are  three  wires,  a  is  2  mils,  I  is  3  mils, 
and  c  is  4  mils  in  diameter;  what  length  must  b 
and  c  have  to  be  equal  in  resistance  to  ten  feet  of  af 

Solution:  Take  a  and  c  first  and  apply  the  rule, 
]0  X  49  -2s  =  40  feet;  then  take  a  and  I  10  X 
3a  •*•  22  =  22^  feet.  To  prove  it  compare  a  and  c 
directly  by  the  same  rule  22*4  X  42  -*-  32  =  40. 
As  this  gives  the  same  result  as  the  first  operation, 
we  may  regard  it  as  proved. 

A  conductor  is  75  mils  in  diameter  and  79  feet 
long;  how  thick  must  a  wire  1264  feet  long  be  to 
equal  it  in  resistance? 

Solution:  752  X  1264  •*•  79  =  7,110,000  -5-  79  = 
90,000.  The  square  root  of  this  amount  is  300  which 
is  the  required  diameter. 


28  ARITHMETIC  OF  ELECTRICITY. 

For  problems  involving  the  comparison  of  wires  of 
unequal  resistance  the  rule  may  be  thus  stated: 

Rule  15.  Multiply  the  square  of  the  diameter  of  each 
'wire  by  the  length  of  the  other.  Of  the  two  products 
divide  the  one  by  the  other  to  get  the  ratio  of  resist- 
ance of  the  dividend  to  that  of  the  divisor  taken  at 
unity.  The  term  including  the  length  of  a  given  wire 
is  the  one  expressing  the  relative  resistance  of  such 
wire. 

EXAMPLES. 

A  wire  is  40  mils  in  diameter,  3  miles  long  and 
40  ohms  resistance.  A  second  wire  is  50  mils  in 
diameter  and  9  miles  long.  What  is  its  resistance? 

Solution:  9  X  402  =  14,400  relative  resistance  of 
the  first  wire.  3  X  502  =  7,500  relative  resistance 
of  second  wire.  14,400  -*-  7,500  =  1.92  —  ratio  of 
resistance  of  second  wire  to  that  of  first  taken  at 
unity.  But  the  latter  resistance  really  is  40  ohms. 
Therefore  the  resistance  of  the  second  wire  is  40  X 
1.92  =  76.80  ohms. 

The  result  may  also  be  worked  out  thus: 

402  X  9  =  14,400  =  relative  resistance  of  the  3  mile 
wire. 

502  x  3  =  7500  =  relative  resistance  of  the  9  mile 
wire. 

14,400  •*•  7500  =  1.92  =  ratio  of  9  mile  (dividend) 
to  3  mile  (divisor)  wire. 

.-.  40  ohms  X  1.92  =  76.8  ohms. 

A  length  of  a  thousand  feet  of  wire  95  mils  in 
diameter  has  1.15  ohms  resistance;  what  is  the  di- 


RESISTANCE  AND  CONDUCTANCE.  29 

ameter  of  a  wire  of  the  same  material  of  which  the 
resistance  of  1000  feet  is  10.09  ohms?  (R.  E.  Day, 
M.  A.). 

Solution:  10.09  •*-  1.15  =  8. 7?  ratio  of  resistances. 
If  we  divide  1000  by  8. 77  we  obtain  a  length  of  the 
first  wire  which  reduces  the  question  to  one  of  iden- 
tical resistances.  1000  •+•  877  =  114  feet.  Then 
applying  Rule  14,  114  X  95«  •*•  1000  =  1037.88. 
This  is  the  square  of  the  diameter  of  the  other  wire. 
Its  square  root  gives  the  answer:  32.2  mils. 

SPECIFIC  RESISTANCE. 

Specific  resistance  is  .the  resistance  of  a  cube  of 
one  centimeter  diameter  of  the  substance  in  ques- 
tion between  opposite  sides.  It  is  expressed  in 
ohms  for  solutions  and  in  microhms  for  metals. 
From  it  may  be  determined  the  resistance  of  all 
volumes,  generally  prisms  or  cylinders,  of  substance. 
Very  full  tables  of  Specific  Resistance  are  given  in 
their  place. 

Rale  16.  The  resistance  of  any  prism  or  cylinder  of  a 
substance  is  equal  to  its  specific  resistance  multiplied 
by  its  length  in  centimeters  and  divided  by  its  cross- 
sectioiial  area  in  square  centimeters.  If  the  dimensions 
are  given  in  inches  or  other  units  of  measurements 
they  must  be  reduced  to  centimeters  by  the  table. 

Sp.  R  x  I 

* 

EXAMPLES. 

An  electro-plater  has  a  bath  of  sulphate  of  copper, 
gp.  resistance  40  ohms.  His  electrodes  are  each  1 


80  ARITHMETIC  OF  ELECTRICITY. 

foot  square  and  1  foot  apart.  What  is  the  resist* 
ance  of  such  a  bath  ? 

Solution :  By  the  table  1  square  foot  =  929  sq.  cent, 
and  1  foot  =  30.4797  cent.  .-.  Resistance  =  40  X 
30.4797  •+•  929  =  1.31  ohms. 

Where  the  electrodes  in  a  solution  are  of  uneven 
size  take  their  average  size  per  area.  The  facing 
areas  are  usually  the  only  ones  calculated,  as  owing 
to  polarization  the  rear  faces  are  of  slight  efficiency, 
and  where  the  electrodes  are  nearly  as  wide  as  the 
bath  or  cell  the  active  prism  is  practically  of  cross- 
sectional  area  equal  to  the  area  of  one  side  of  a  plate. 

In  a  Bunsen  battery  the  specific  resistances  of  the 
solutions  in  inner  and  outer  cells  were  made  alike, 
each  equalling  9  ohms.  The  central  element  was  a 
YZ  inch  cylinder  of  electric  light  carbon.  The  outer 
element  was  a  plate  of  zinc  6  inches  long  bent  into  a 
circle.  When  there  were  2  inches  of  solution  in  the 
cell  what  was  the  resistance? 

Solution  :  Area  of  carbon  =  f  X  2  =  3.14  square 
inches.  Area  of  zinc  =  2  X  6  =  12  square  inches. 
This  gives  an  average  facing  area  of  (12  +  3.14)  -*- 
2  =  7.57  square  inches  =  48.38  sq.  cent.  The 
distance  apart  =  ^  inches  (nearly)  =  1.9049  cent. 
.-.  Resistance  =  9  X  1.9049  -*-  48.38  =  .354  ohms. 

For  wires,  the  specific  resistance  of  metals  being 
given  in  microhms,  the  calculation  may  be  made  in 
microhms,  or  in  ohms  directly.  As  wire  is  cylindri- 


RESISTANCE  AND  CONDUCTANCE.  3] 

cal  a  special  calculation  may  be  made  in  its  case  to 
reduce  area  of  cross  section  to  diameter.  This  may 
readily  be  taken  from  the  table  of  wire  factors,  thus 
avoiding  all  calculation. 

Rule  17.  The  resistance  In  microhms  of  a  wire  of  given 
diameter  In  centimeters  Is  equal  to  tlie  product  of  the 
specific  resistance  by  1.2737  by  the  length  In  centi- 
meters divided  by  the  square  of  the  diameter  In  cen- 
timeters. 

«  _  Sp.  Res,  x  '.2737  x  L 
d» 

EXAMPLES. 

The  Sp.  Res.  of  copper  being  taken  at  1.652  mi- 
crohms what  is  the  resistance  of  a  meter  and  a  half 
of  copper  wire  1  millimeter  thick? 

Solution:  The  diameter  of  the  wire  (1  millimeter) 
is  .1  centimeter.  The  square  of  .1  is  .01.  The 
length  of  the  wire  (iy2  meter)  is  150  centimeters. 
Its  resistance  therefore  is  1.652  X  1.2737  X  150  -H 
.01  =  31,561  microhms  or  .031,561  ohms. 

UNIVERSAL  RULE  FOR  RESISTANCES. 

Into  the  problem  of  resistances  of  one  or  two  wires 
eight  factors  can  enter,  these  are  the  lengths,  sec- 
tional areas,  specific  resistances  and  absolute  resist- 
ances of  two  wires.  Their  relation  may  be  ex- 
pressed by  an  algebraic  equation,  which  by  transposi- 
tion may  be  made  to  fit  any  case.  The  rule  is 
arithmetically  expressed  by  adopting  the  method  of 
cancellation,  drawing  a  vertical  line  and  placing  on 


32  ARITHMETIC  OF  ELECTRICITY. 

the  left  side,  factors  to  be  multiplied  together  for 
a  divisor,  and  on  the  right  side  factors  to  be  multi- 
plied together  for  a  dividend.  In  the  expression  of 
the  rule  as  below  the  quotient  is  1,  in  other  words 
the  product  of  all  the  factors  on  the  left  hand  of 
the  line  is  equal  to  that  of  all  the  factors  on  the 
right  hand.  Calling  one  wire  a  and  the  other  b  we 
have  the  following  expression: 


Resistance  of  1) 
Specific  Resistance  of  a 
Length  of  a 
Cross-sectional  area  of  b 


Resistance  of  a 
Specific  Resistance  of  b 
Length  of  b 
Cross-sectional  area  of  d 


Rule  18.  Substitute  in  the  above  expression  the  values 
of  any  factors  given.  Substitute  for  factors  not  given 
or  required  the  figure  1  or  unity.  Such  a  value  deter- 
mined by  division  must  be  given  to  the  required  factor 
and  substituted  in  its  place  as  will  make  the  product 
of  the  left-hand  factors  equal  to  that  of  the  right-hand 
factors.  Only  one  factor  can  be  determined,  and  all 
factors  not  given  are  assumed  to  be  respectively  equal 
for  both  conductors. 

EXAMPLES. 

If  the  resistance  of  500  feet  of  a  certain  wire  is 
009  ohms  what  is  the  resistance  of  1050  feet  of  the 
same  wire? 

Solution:  The  cross  sectional  areas  and  specific 
resistance  not  being  given  are  taken  as  equal.  (This 
of  course  follows  from  the  identical  wire  being  re- 
ferred to.)  The  vertical  line  is  drawn  and  the 
values  substituted  : 


RESISTANCE  AND  CONDUCTANCE.  33 

Resistances :     Eesistance  of     .09 

required  wire 
Lengths :  500     1050 

(Other  factors  omitted  as  unnecessary.) 
1050  X  .09  •*•  500  =  .189  ohms. 

What  is  the  diameter  of  a  wire  2  miles  long  of 
23  ohms  resistance,  if  a  mile  of  wire  of  similar  ma- 
terial of  seventy  mils  diameter  has  a  resistance  of 
10. 82  ohms? 

Solution.  We  use  for  simplicity  the  square  of  the 
diameter  in  place  of  the  cross  sectional  area  of  the 
known  wire,  thus: 


Resistances :  23 

Lengths :  1 

Areas :  Unknown 


10.82 

2 

702 


As  the  specific  resistances  are  identical  they  are 
not  given. 

2  X  ?02  X  10.82  •*•  23  X  1  =  4610  square  of  diameter 
required :  4610^  =  68  mils. 

What  must  be  the  length  of  an  iron  wire  of  cross- 
sectional  area  4  square  millimeters  to  have  the  same 
resistance  as  a  wire  of  pure  copper  1000  yards  long, 
of  cross-sectional  area  1  square  millimeter,  taking 
the  conductance  of  iron  as  }  that  of  copper?  (Day). 

Solution: 


34  ARITHMETIC  OF  ELECTRICITY. 


Specific  Resistances :      1 

Lengths :  1000 

Cross-sectional  areas :    4 


7  (i.e.  the  reciprocal  of 

conductance) 
Unknown 
1 


As  the  resistances  are  identical  they  are  not 
given. 

Solving  we  have  1000  X  4  •*•  7  =  571f  yards. 

There  are  two  conductors,  one  of  35  ohms  resist- 
ance, 1728  feet  long  and  12  square  millimetres  cross- 
sectional  area  and  specific  resistance  7:  the  other  of 
14  ohms  resistance,  432  feet  long  and  8  square  milli- 
metres cross-sectional  area.  What  is  its  specific 
resistance? 


Resistances :  35 

Specific  Resistances 
Unknown 

Lengths:  432 

Cross- Sectional  areas :  12 


14 


7 

1728 

8 


By  cancellation  this  reduces  to  14X8-*-5X3  — 

7.4  Specific  Resistance. 

In  these  cases  it  is  well  to  call  one  wire  a  and  the 
other  #,  and  to  arrange  the  given  factors  in  two 
columns  headed  by  these  designations.  Then  the 
formula  can  be  applied  with  less  chance  of  error. 
Thus  for  the  last  two  problems  the  columns  should 
be  thus  arranged. 


RESISTANCE  AND  CONDUCTANCE. 


35 


a 

b 

a 

6 

Area.  4  sq.  mils. 
L.  Unknown 
Sp.  Res.  7 

1 

1,000  yards 
1 

Resist.  35  olims 
L.  1,728  feet 
Area,  12  sq.  mils. 
Sp.  Res  7 

14  ohms 
432  feet 
8  sq.  mils, 
required 

From  such  statements  of  known  data  the  formula 
can  be  conveniently  filled  up. 

RESISTANCE  OF  WIRES  REFERRED  TO  WEIGHT. 

The  weight  of  equal  lengths  of  wire  is  in  propor- 
tion to  their  sections.  The  problems  involving 
weight  therefore  can  be  reduced  to  the  Rules  al- 
ready given. 

Problem.  A  wire,  A,  is  334  feet  long  and  weighs  25 
oz.;  another,  B,  is  20  feet  long  and  weighs  1  oz. 
what  are  the  relative  resistances? 

Solution:  20  feet  of  the  wire  "  A"  weigh  ^  x 
25  =  1.50  oz.  The  weights  of  equal  lengths  of  A 
and  B  respectively  areas  1.50  :  1.00  which  is  also 
the  inverse  ratio  of  the  resistances  of  equal  lengths. 
By  compound  proportion  Rule  we  have  R.  of  "  A  " 
:  R.  of  "B"  ::  1  X  334  ::  1.50  X  20;  reducing  to 
16.7  :  1.5  or  11.1  :  1.0  or  the  wire  "  A  "  has  about 
eleven  times  the  resistance  of  the  wire  "  B." 

Solution:  By  general  Rule  for  resistance  (Rule 
18).  Taking  1.50  :  1.00  as  the  ratio  of  cross-sec- 
tional areas  and  taking  the  resistance  of  the  long 
wire  A  as  1  we  have : 


36  ARITHMETIC  OF  ELECTRICITY. 


Resistances :  1 

Lengths :  20 

Cross-sectional  area :  1.50 


Unknown 

334 

1 


Eesistance  of  B  =  1.50  X  20  •*•  334  =  .0899  01 
about  fr  as  before. 

CONDUCTANCE. 

Conductance  is  the  reciprocal  of  resistance  and  is 
sometimes  expressed  in  units  called  MHOS,  which 
is  derived  from  the  word  ohm  written  backwards. 

Rule  19.  To  reduce  resistance  in  ohms  to  conductance 
in  mhos  express  its  reciprocal  and  the  reverse. 

K=R 

EXAMPLES. 

A  wire  has  a  resistance  of  t1^  ohms,  what  is  its 
conductance? 

Solution:  126  -*-  18  =  7  mhos. 

Eeduce  a  conductance  of  lit  to  ohms. 
Solution:  lit  =  If  mhos  which  gives  if  ohms. 

It  is  evident  that  the  data  for  problems  or  that 
constants  could  be  given  in  mhos  instead  of  ohms. 
In  some  ways  it  is  to  be  regretted  that  the  positive 
quality  of  conductance  was  not  adopted  at  the  out- 
set instead  of  the  negative  quality  of  resistance. 
One  or  two  illustrations  may  be  given  in  the  form  of 
examples  involving  conductance. 

Express  Ohm's  law  in  its  three  first  forms  in 
conductance. 


RESISTANCE  AND  CONDUCTANCE.  37 

Solution:  This  is  done  by  replacing  the  factor 
"resistance"  by  its  reciprocal.  Thus,  Rule  1 
reads  for  conductance:  "The  current  is  equal  to 
the  electromotive  force  multiplied  by  the  conduc- 
tance" (C  =  EK) — Rule  2  as  "The  electromotive 
force  is  equal  to  the  current  divided  by  the  conduc- 

C 
tance  "  (E  =  — ) — Rule  3  as  "  The  conductance  is 

K 
equal  to  the  current  divided  by  the  electromotive 

C 
force."  (K  =  — ) 

E 

A  circuit  has  a  resistance  of  .5  ohm  and  an  E.  M. 
F.  of  50  volts;  determine  the  current,  using  con- 
ductance method. 

Solution:  The  conductance  =  .-J-  =  2  mhos.  The 
current  =  50  X  2  =  100  amperes. 

In  a  circuit  a  current  of  20  amperes  is  main- 
tained through  2f  ohms.  Determine  the  E.  M.  F. 
using  conductance. 

Solution:    The  conductance    =  4    =    tt    mhos. 

E.  M.  F  =  20  •*•  H  =  52  volts. 

Assume  a  current   of  30  amperes  and  an  E.  M. 

F.  of  50  volts,  what  is  the  conductance  and  resis- 
tance? 

Solution:  Conductance  =  30  •*•  50  =  .6  mho. 
Resistance  =  1  -*•  .6  =  1.667. 


CHAPTER  IV. 

POTENTIAL  DIFFERENCE. 

DROP   OF   POTENTIAL   IN    LEADS    AND    SIZE    OF 
SAME  FOR  MULTIPLE  ARC  CONNECTIONS. 

SUBSIDIARY  leads  are  leads  taken  from  large  sized 
mains  of  constant  E.  M.  F.  or  from  terminals 
of  constant  E.  M.  F.  to  supply  one  or  more  lamps, 
motors,  or  other  appliances.  A  constant  voltage  is 
maintained  in  the  mains  or  terminals.  There  is  a 
drop  of  potential  in  the  leads  so  that  the  appliances 
always  have  to  work  at  a  diminished  E.  M.  F.  The 
E.  M.  F.  of  the  leads  is  known,  the  requisite 
E.  M.  F.  and  resistance  of  the  appliance  is  known, 
a  rule  is  required  to  calculate  the  size  of  the  wire  to 
secure  the  proper  results.  It  is  based  on  the  princi- 
ple that  the  drop  or  fall  in  potential  in  portions  of 
integral  circuits  varies  with  the  resistance.  (See 
Ohm's  law).  A  rule  is  required  for  a  single  appli- 
ance or  for  several  connected  in  parallel. 

Rule  20.  The  resistance  of  the  leads  supplying  any  ap- 
pliance or  appliances  for  a  desired  drop  in  potential 
within  the  leads  is  equal  to  the  reciprocal  of  the  cur- 
rent  of  the  appliances  multiplied  by  the  desired  drop 
in  volts. 


POTENTIAL  DIFFERENCE.  39 

EXAMPLE. 

A  lamp,  100  volts  X  200  ohms,  is  placed  100  feet 
from  the  mains,  in  which  mains  a  constant  E.  M.  F. 
of  110  volts  is  maintained.  What  must  be  the  resist- 
ance of  the  line  per  foot  of  its  length;  and  what 
size  copper  wire  must  be  used? 

Solution:  The  lamp  current  is  obtained  (Ohm's 
law)  by  dividing  its  voltage  by  its  resistance,  (&$  = 
\  ampere).  The  reciprocal  of  the  current  is  f;  mul- 
tiplied by  the  drop  (f  X  10  =  20)  it  gives  the  resist- 
ance of  the  line  as  20  ohms.  As  the  lamp  is  100 
feet  from  the  mains  there  are  200  feet  of  the  wire. 
Its  resistance  per  foot  is  therefore  ^  =  -fa  ohm  or 
it  is  No.  30  A.  W.  G.  (about). 

For  several  appliances  in  parallel  on  two  leads  a 
similar  rule  may  be  applied.  There  is  inevitably  a 
variation  in  E.  M.  F.  supplied  to  the  different  appli- 
ances unless  resistances  are  intercalated  between  the 
appliances  and  the  leads. 

Rule  21.  The  K.  in.  F.  of  the  main  leads  or  terminals 
the  factors  of  the  lamps  or  other  appliances,  tbeir  num. 
ber  and  the  distance  of  their  point  of  connection  are 
given.  The  combined  resistance  is  found  by  Rules  8  to 
12.  Then  by  Rule  20  the  resistance  of  the  lead*  is  cal- 
culated. 

EXAMPLE. 

A  pair  of  house  leads  includes  260  feet  of  wire, 
or  130  in  each  lead.  Six  50  volt  100  ohm  lamps  are 
connected  thereto  at  the  ends.  The  drop  is  to  be  5 


40  ARITHMETIC  OF  ELECTRICITY. 

volts,  giving  55  volts  in  the  main  leads.  Eequired  the 
total  resistance  of  and  size  of  wire  for  the  house  leads. 
Solution:  The  resistance  of  six  100  ohm  lamps  in 
parallel  is  100  -s-  6  =  16.66  ohms.  The  current  re- 
quired is  by  Ohm's  law  50  •*-  16.66  or  3  amperes. 
Its  reciprocal  multiplied  by  the  drop,  (1X5=1  = 
l^i  ohms)  gives  the  required  resistance  =  IJ/s  ohms. 
This,  divided  by  260  feet  gives  the  resistance  per 
foot  as  .0064  ohm,  corresponding  by  the  table  to 
No.  18  A.  W.  G. 

A  rule  for  the  above  cases  is  sometimes  expressed 
otherwise,  being  based  on  the  proportion:  Eesist- 
ance  of  appliances  is  to  resistance  of  leads  as  100 
minus  the  drop  expressed  as  a  percentage  is  to  the 
drop  expressed  as  a  percentage.  This  gives  the  fol- 
lowing: 

Rule  22.  The  resistance  of  the  leads  is  equal  to  the 
combined  resistance  of  the  appliances  multiplied  by 
the  percentage  of  drop  and  divided  by  100  minus  the 
percentage  of  drop. 

Problem.  Take  the  data  of  last  problem  and 
solve. 

Solution:  The  percentage  of  drop  is  &  =  9$.  The 
resistance  of  the  leads  =  16-66  x  9  =  149.94  =  \yz  Ohms 

100  —  9  91 

about. 

Note. — To  obtain  accurate  results  the  figures  of  percen- 
tage, etc.,  must  be  carried  out  to  two  or  more  decimal 
places.  Rules  20  and  21  are  to  be  preferred  to  any  percen- 
tage rule.  Also  see  Rule  23. 


POTENTIAL  DIFFEEENCE.  41 

Where  groups  of  lamps  are  to  be  connected  along 
a  pair  of  leads  but  at  considerable  intervals,  the 
succeeding  sections  of  leads  have  to  be  of  diminish- 
ing size.  The  same  problem  arises  in  calculating 
the  sizes  of  street  leads.  The  identical  rule  is  ap- 
plied, care  being  taken  to  express  correctly  the  ex- 
act current  going  through  each  section  of  the  lead. 
The  calculation  is  begun  at  the  outer  end  of  the 
leads.  A  diagram  is  very  convenient;  it  may  be 
conventional  as  shown  below. 

EXAMPLES. 

At  three  points  on  a  pair  of  mains  three  groups  of 
fifty  220  ohm  lamps  in  parallel  are  connected;  a 
total  drop  of  5  volts  is  to  be  divided  among  the 
three  groups,  thus:  1.6  volts  —  1.6  volts  —  1.8  volts-. 
The  initial  E.  M.  F.  is  115  volts;  what  must  be  the 
resistances  of  the  three  sections  of  wire? 

Solution:  The  following  diagram  gives  the  data 
as  detailed  above: 

1.  2.  3. 


Starting  at  group  3  we  have  50  lamps  in  parallel 
each  of  220  ohms  resistance,  giving  a  combined  re- 
sistance (Rule  12)  of  4.4  ohms  and  a  total  current 
(Ohm's  law)  of  110  -*-  4.4  =  25  amperes.  The  re- 
sistance of  section  2 — 3  is  by  the  present  rule  $5  X 


42  ARITHMETIC  OF  ELECTRICITY. 

1.8  =  .072  ohms.  Taking  group  2  the  current 
through  this  group  of  lamps  is  111.8  •*•  4.4  =  25.41 
amperes.  The  section  1 — 2  has  to  pass  also  the  cur- 
rent 25  amperes  for  group  3  giving  a  total  current 
of  25  +  25.41  =  50.41  amperes.  The  resistance  of 
section  1 — 2  is  therefore^  X  1.6  =  .0317  ohm. 
Taking  group  1  the  current  for  its  lamps  is  113.4  •*- 
4.4  =  25.7  amperes.  The  total  current  through 
section  0—1  is  therefore  25  +  25.4  +  25.7=  76.1 
amperes.  The  resistance  of  the  section  is  ^  X  1.6 
=  .021  ohms.  Arranging  all  these  data  upon  a 
diagram  we  have  the  full  presentation  of  the  calcu- 
lation in  brief  as  below: 


r 


CHAPTER  V. 

CIRCULAR     MILS. 

A  MIL  is  ntar  of  an  inch.  The  area  of  a  circle, 
one  mil  in  diameter,  is  termed  a  circular  mil.  The 
area  of  the  cross-section  of  wires  is  often  expressed  in 
circular  mils.  Thus  a  wire,  3  mils  in  diameter,  has 
an  area  of  9  circular  mils,  as  shown  in  the  cut.  A 


circular  mil  is  .7854  square  mil.  Rules  for  the  sizes 
of  wires  for  given  resistances  are  often  based  on  cir- 
cular mils,  and  include  a  constant  for  the  conduc- 
tivity of  copper.  By  the  table  of  specific  resistances, 
the  values  found  can  be  reduced  to  wires  of  iron  or 
other  metals. 


44  ARITHMETIC  OF  ELECTRICITY. 

A  commercial  copper  wire,  one  foot  long,  and  one 
circular  mil  in  section,  has  a  resistance  of  10.79 
ohms  at  75°  F.  This  is,  of  course,  largely  an  as- 
sumption, but  is  taken  as  representing  a  good  aver- 
age. No  two  samples  of  wire  are  exactly  alike,  and 
many  vary  largely.  From  Rule  13,  and  from  the 
above  constant,  we  derive  the  following  rules: 

Rule  23.  The  resistance  of  a  commercial  copper  wire 
is  equal  to  its  length  divided  by  the  cross-section  in  cir- 
cular mils,  and  multiplied  by  10.79. 

EXAMPLE. 

A  wire  is  1050  feet  long,  and  has  a  cross-section 
of  8234  circular  mils.     What  is  its  resistance? 
Solution:  1050  X  10.79  •*-  8234  =  1.37'  ohms. 

Rule  24.  The  cross-section  of  a  wire  in  circular  mils 
is  equal  to  its  length  divided  by  its  resistance,  and  mul- 
tiplied by  10.79. 

EXAMPLE. 

A  wire  is  1050  feet  long,  and  has  a  resistance  of 
.68795  ohms.  What  is  its  cross-section  in  circular 
mils? 

Solution:  1050  X  10.79  -*-  .68795  =  16,468  circu- 
lar mils. 

Rule  25.  The  cross-section  of  the  wire*  of  a  pair  of 
leads  in  circular  mils  for  a  given  drop  expressed  in  per- 
centage is  equal  to  the  product  of  the  length  of  leads  by 
the  number  of  lamps  (in  parallel),  by  21.58,  by  the  dif- 
ference between  1OO  and  the  drop,  the  whole  divided 
by  the  resistance  of  a  single  lamp  multiplied  by  the 
drop.  __  Inx  21.58  X  (100-  e) 

A  ~~  er 


CIRCULAR  MILS.  45 

EXAMPLE. 

Fifty  lamps  are  to  be  placed  at  the  end  of  a  double 
lead  150  feet  long  (=  300  feet  of  wire).  The  resist- 
ance of  a  lamp  is  220  ohms.  What  size  must  the 
wire  be  for  5%  drop? 

Solution:  150  X  50  X  21.58  X  (100  —  5)  -*-  (220  X 
5)  =  13,977.9  circular  mils. 

In  these  calculations  and  in  the  calculations  given 
on  page  48  it  is  important  to  bear  in  mind  that  the 
percentage  is  based  upon  the  difference  of  potential 
at  the  beginning  of  the  leads  or  portion  thereof 
under  consideration;  in  other  words  upon  the  high- 
est difference  of  potential  within  the  system  or  the 
portion  of  the  system  treated  in  the  calculation. 


CHAPTER  VI. 

SPECIAL    SYSTEMS. 

THKEE  WIRE  SYSTEM. 

As  there  are  three  wires  in  the  three  wire  system, 
where  there  are  two  in  the  ordinary  system,  and  as 
each  of  the  three  wires  is  one  quarter  the  size  of 
each  of  the  two  ordinary  system  wires,  the  copper 
used  in  the  three  wire  system  is  three-eighths  of 
that  used  in  the  ordinary  system. 

In  the  three  wire  system  the  lamps  are  arranged 
in  sets  of  two  in  series.  Hence  but  one-half  the 
current  is  required.  The  outer  wires,  it  will  be  no- 
ticed, have  double  the  potential  of  the  lamps. 
Hence  to  carry  one-half  the  current  with  double  the 
E.  M.  P.,  a  wire  of  one  quarter  the  size  used  in  the 
ordinary  system  suffices. 

Rule  26.  The  calculations  for  plain  multiple  arc 
work  apply  to  the  three  wire  system,  as  regards  size  of 
each  of  the  three  leads,  If  divided  by  4. 

While  the  central  or  neutral  wire  will  haye  noth- 
ing to  do  when  an  even  number  of  lamps  are  burning 
on  each  side,  and  may  never  be  worked  to  its  full 
capacity,  there  is  always  a  chance  of  its  having  to 
carry  a  full  current  to  supply  half  the  lamps  (all  on 


SPECIAL  SYSTEMS.  47 

one  side).     Hence  it  is  made  equal  in  size  to  the 
others. 

ALTERNATING  CURRENT  SYSTEM. 

The  rules  already  given  apply  in  practice  to  this 
system  also,  although  theoretically  Ohm's  law 
and  those  deduced  from  it  are  not  correct  for  this 
current.  A  calculation  has  to  be  made  to  allow  for 
the  conversion  from  primary  to  secondary  current 
in  the  converter  as  below. 

Note. — The  ratio  of  primary  E.  M.  F.  to  secondary  is  ex- 
pressed by  dividing  the  primary  by  the  secondary,  and  is 
termed  ratio  of  conversion.  Thus  in  a  1000  volt  system  with 
50  volt  lamps  in  parallel  the  ratio  of  conversion  is  1000  -*-  50 
=  20. 

Rule  27.  The  current  in  tbe  primary  is  eqnal  to  the 
current  in  the  secondary  divided  by  the  ratio  of  conver- 
sion. 

EXAMPLES. 

On  an  alternating  current  circuit  whose  ratio  of 
conversion  =  20,  there  are  1000  lamps,  each  50  volt; 
50  ohms.  When  all  are  lighted  what  is  the  primary 
current  ? 

Solution :  By  Ohm's  law  the  secondary  current  is 
1000  x  1  (each  lamp  using  -f$  =  1  ampere,  Ohm's 
law)  =  1000  amperes.  1000  -«-•  20  =  50  amperes  is 
the  primary  current. 

The  current  being  thus  determined  the  ordinary 
rules  all  apply  exactly  as  given  for  direct  current 
work. 


48  ARITHMETIC  OF  ELECTRICITY. 

Given  650  lamps,  50  volt  50  ohms  each,  3600 
feet  from  the  station.  The  primary  circuit  pressure 
is  1031  volts.  A  drop  of  3$  is  to  be  allowed  for  in 
the  primary  leads.  What  is  the  resistance  of  the 
primary  wire? 

Solution:  Current  of  a  single  lamp  =  50  -*-  50  =  1 
ampere;  current  of  650  lamps  =  650  amperes,  cur- 
rent of  primary  650  -*-  20  =  32£  amperes,  drop  of 
primary  =  1031  X  3#  =  30.9  volts,  resistance  of  pri- 
mary (Rule  20)  £  X  30.9  =  .9516  ohm. 

Rule  28.  For  obtaining  tlie  size  of  tlie  primary  wire 
In  circular  mils,  calculate  by  Rule  25,  and  divide  the 
result  by  the  square  of  the  ratio  of  conversion. 

EXAMPLE. 

Take  data  of  last  problem  and  solve. 
Solution:  [3600  X  650  X  21.58  X  (100—3)  •*-  (50  X 
3)]  -*•  202  =  81,637  circular  mils. 

The  two  last  examples  may  be  made  to  prove  each 
other,  thus: 

The  total  length  of  leads  is  3600  X  2  ==  7200  feet. 
If  of  1  mil  thickness  its  resistance  would  be  7200  X 
10.79  =  77,688  ohms.  As  resistance  varies  inversely 
as  the  cross  sectional  area  we  have  the  proportion 

.9516  :  77,688  ::  1  :  x  which  gives  x  =  81,639  cir- 
cular mils  corresponding  within  limits  to  the  result 
obtained  by  Rule  28. 

In  all  cases  of  this  sort  where  percentage  is  ex- 
pressed the  statement  in  the  last  paragraph  on  page 


SPECIAL  SYSTEMS.  49 

45  should  be  kept  in  mind.  The  ratio  of  conversion 
must  be  based  on  the  E.  M.  F.  at  the  coil  (in  this 
case  on  1031  —  31  =  1000  volts)  not  on  the  E.  M.  F. 
at  the  beginning  of  the  leads  or  portion  thereof  con- 
sidered in  the  calculation.  The  percentage  of  drop 
must  be  subtracted  before  the  ratio  of  conversion  is 
calculated. 

For  winding  the  converters,  the  following  is  the 
rule  : 

Rule  29.  The  convolutions  of  the  primary  are  equal 
in  number  to  the  product  of  the  convolutions  of  the  sec- 
ondary multiplied  by  the  ratio  of  conversion,  and  vice 
versa. 

EXAMPLES. 

The  ratio  of  conversion  of  a  coil  is  20;  there  are 
1000  convolutions  of  th,e  secondary.  How  many 
should  there  be  of  the  primary? 

Solution:  1000  X  20  =  20,000  convolutions. 

There  are  in  a  coil  5000  convolutions  of  the  pri- 
mary; its  ratio  of  conversion  is  50.  How  many  con- 
volutions should  the  secondary  have? 

Solution:  5000  •*-  50  =  100  convolutions. 


CHAPTER  VIL 

WORK   AND    ENERGY. 

ENERGY  AND  HEATING  EFFECT  OF  THE  CURRENT. 

IT  has  been  shown  experimentally  by  Joule  that 
the  total  quantity  of  heat  developed  in  a  circuit  is 
equal  to  the  square  of  the  current  multiplied  by  the 
resistance.  This  is  equal,  by  Ohm's  law,  to  the 
square  of  the  E.  M.  F.  divided  by  the  resistance, 
which  again  reduces  to  the  E.  M.  F.  multiplied  by 
the  current.  Each  of  these  expressions  has  its  own 
application,  and  they  may  be  given  as  three  rules. 

Rule  30.  The  energy  or  neat  developed  in  "circuits  is 
in  proportion  to  the  square  of  the  current  multiplied  bjr 
the  resistance. 


EXAMPLES. 

An  electric  lamp  has  a  resistance  of  50  ohms;  it  is 
connected  to  a  street  main  by  leads  of  21  ohms  re- 
sistance. What  proportion  of  heat  is  wasted  in  the 
house  circuit? 

Solution:  The  current  being  the  same  for  all  parts 
of  the  circuit,  the  heat  developed  is  in  proportion  to 
the  resistance,  or  as  2i  :  50  equal  to  1  :  20.  The 


WORK  AND  ENERGY.  51 

heat  developed  in  the  wire  is  wasted,  therefore  the 
waste  is  £>  of  the  total  heat  developed. 

The  internal  resistance  of  a  battery  is  equal  to 
that  of  3  meters  of  a  particular  wire.  Compare  the 
quantities  of  heat  produced  both  inside  and  outside 
the  battery  when  its  poles  are  connected  with  one 
meter  of  this  wire  with  the  quantities  produced  in 
the  same  time  when  they  are  connected  by  37  meters 
of  the  same  wire.  (Day.) 

Solution:  The  relative  currents  produced  in  the 
two  cases  are  found  (Ohm's  law)  by  dividing  the 
E.  M.  F.  of  the  battery  (a  constant  quantity  =  E) 
by  the  relative  resistance.  As  the  battery  counts 
for  the  resistance  of  3  meters  of  wire,  the  relative 
resistances  are  as  4  :  40.  "Were  the  same  current  de- 
veloped in  both  cases  these  figures  would  give  the 
desired  ratio.  But  as  the  current  varies  it  has  to  be 
taken  into  account.  To  determine  the  relative  bat- 
tery heat  only,  we  may  neglect  resistance  of  the  bat- 
tery, as  it  is  a  constant  for  both  cases,  the  battery 
remaining  identical.  By  Ohm's  law  the  currents 
are  in  the  ratio  of  f  :  ^  and  their  squares  are  in  pro- 
portion to  f|  :  T|J7  =  100  :  1.  By  the  rule  this  is 
the  proportion  of  the  heats  developed  in  the  battery 
alone,  with  the  short  wire  100,  with  the  long  wire 
1.  "For  the  heating  effects  on  the  outside  circuit, 
as  resistance  and  current  both  vary,  the  full  for- 
mula of  the  rule  must  be  applied.  The  ratio  of  the 
heat  in  the  short  wire  connection  to  that  in  the  long 


52  ARITHMETIC  OF  ELECTRICITY. 

wire  connection  is  as  (f)2  X  1  :  (^)2  X  37  =  100  X  1 : 
37  X  1.  The  ratio  therefore  is  as  100  is  to  37  for 
the  total  heat  produced  in  the  circuit  which  includes 
battery  and  connections. 

Owing  to  irregular  working  of  a  dynamo,  an  in- 
candescent lamp  receives  sometimes  the  full  amount 
or  i  ampere  of  current;  at  other  times  as  little  as  i$ 
ampere.  What  proportion  of  heat  is  developed  in  it 
in  both  cases,  assuming  its  resistance  to  remain 
sensibly  the  same? 

Solution:  By  the  rule  the  ratio  is  (i)a :  ($f)  or  J: 
^A°T  =  2025  :  400.  The  diminution  of  current  there- 
fore cuts  down  the  heat  to  $  the  proper  amount. 

Rule  31.  The  energy  or  heat  developed  in  a  circuit  is 
in  proportion  to  the  square  of  the  electromotive  force 
divided  by  the  resistance.  E" 

~B 
EXAMPLES. 

There  are  two  Grove  batteries,  each  developing 
1.98  volts  E.  M.  F.  One  has  an  internal  resistance 
of  -A-  ohm;  the  other  of  *  ohm.  They  are  placed  in 
succession  on  a  circuit  of  2  ohms  resistance.  What 
is  the  ratio  of  heats  developed  by  the  batteries  in 
each  case? 

Solution:  As  the  E.  M.  F.  is  constant  it  may  be 
taken  as  unity.  Then  for  the  two  cases  we  have 
2A>  *  24  ssa  ^i  •  %ds  as  the  ratio  of  heat  produced. 

A  battery  of  one  ohm  resistance  and  two  volts  E. 


WORK  AND  ENERGY.  53 

M.  F.  is  put  in  circuit  with  4  ohms  resistance. 
Another  battery  of  4  ohms  and  1  volt  is  connected 
through  1  ohm  resistance.  What  ratio  of  heat  is 
developed  in  each  case? 

Solution:  i|_2  :  1|1  or  4  :  1. 

Kule  32.  The  energy  or  heat  developed  In  a  circuit  I» 
in  proportion  to  the  E.  31.  F.  multiplied  by  the  current. 

H  =  Ef 

EXAMPLES. 

Take  data  of  last  problem  and  solve. 

Solution:  For  first  battery,  by  Ohm's  law,  current  = 
£  ampere;  for  the  second,  current  =  $  ampere.  The 
heat  developed,  is  by  the  present  rule,  in  the  propor- 
tion as  |  X  2  :  £  X  1  or  4  :  1. 

Compare  the  heat  developed  in  a  100  volt  200  ohm 
lamp  and  in  a  35  volt  35  ohm  lamp  and  in  a  50  volt 
50  ohm  lamp. 

Solution  :'The  currents  (Ohm's  law)  are  :  M&M  and 
!§  in  amperes  =  I,  1,  and  1  amperes.  The  heats  devel- 
oped are,  by  the  rule,  in  the  ratio  100  X  |  :  35  X  1 
and  50  X  1  or  50  :  35  :  50. 

The  same  problem  can  be  done  directly  by  Rule 
31,  thus:  The  three  lamps  develop  heat  in  the 
ratio  m*  :  &  :  3$'  =  50  :  35  :  50.  This  is  the  direct 
and  preferable  method  of  calculation. 


,—  For  "  heat,"  "rate  of  energy,"  or  "rate  of  work" 
can  be  read  in  these  rules. 


54  ARITHMETIC  OF  ELECTRICITY. 

THE  JOULE  OR  GRAM-CALORIE. 

The  last  rules  and  problems  only  touch  upon  the 
relative  proportions  of  heat;  they  do  not  give  any 
actual  quantity.  If  we  can  express  in  units  of  the 
same  class  a  standard  quantity  of  heat,  then  by  deter- 
mining the  relation  of  the  standard  to  any  other 
quantity,  we  arrive  at  a  real  quantity.  Such  a  stan- 
dard is  the  joule,  sometimes  called  the  "  calorie"  or 
"gram-calorie."  A  joule  is  the  quantity  of  heat 
required  to  raise  the  temperature  of  1  gram  of 
water  1  degree  centigrade.  It  is  often  expressed 
as  a  water-gram-degree  0.  or  w.  g.  d.  0.  or  for 
shortness  g.  d.  C.,  from  the  initials  of  the  factors. 
It  is  unfortunate  that  it  is  called  the  calorie 
as  the  name  is  common  to  the  water-kilogram- 
degree  C.  or  kg.d.  0.  The  joule  is  equal  to  4.16 
X  107  or  41,600,000  ergs. 

It  will  be  remembered  that  practical  electric  units 
are  based  on  multiples  of  the  C.  G.  S.  units  of  which 
the  erg  is  one.  The  joule  comes  in  the  C.  G.  S. 
order.  Therefore  to  determine  quantities  of  heat 
the  following  is  a  general  rule  when  the  practical 
electric  units  are  used. 

Rule  33.  Obtain  the  expression  of  rate  of  heat  devel- 
oped, or  of  rate  of  energy,  or  of  rate  of  work  In  volt 
amperes.  Reduce  to  C.  G.  S.  units  (ergs)  by  multiply- 
ing by  1OT  and  divide  by  the  value  of  a  joule  in  ergs 
(4.1 6  X  1O7).  The  quotient  is  joules  or  water-gram  de- 
grees C.  per  second. 

Q  _  E  x  C  X  107 
*•  ~    4.16  X  107 


WORK  AND  ENERGY.  55 

EXAMPLE. 

A  current  of  20  amperes  is  flowing  through  a  wire. 
What  heat  is  developed  in  a  section  of  the  wire 
whose  ends  differ  in  potential  by  110  volts? 

Solution:  The  rate  of  energy  in  watts  or  volt- 
amperes  =  110  X  20  =  2200.  In  the  C.  G-.  S.  units 
this  is  expressed  by  (110  X  108)  X  (20  X  10"1)  =  2200 
X  107  ergs,  per  second;  .  •.  quantity  of  heat  =  2200  X 
107  -s-  4.16  X  107  =  528.8  joules  or  gram-degrees- 
centigrade  per  second. 

As  107  •*-  107  =  1  the  rule  can  be  more  simply 
stated  thus: 

Rule  34.  The  quantity  of  heat  produced  per  second 
in  a  circuit  by  a  current  is  equal  to  the  product  of  the 

watts  by  ^^g  or  by  .24. 

Q,  =  0.24  CE.  or  0.24 1? 

EXAMPLES. 

A  difference  of  potential  of  5.5  volts  is  main- 
tained at  the  terminals  of  a  wire  of  £>  ohm  resist- 
ance. How  many  joules  per  second  are  developed? 

Solution:  By  Ohm's  law,  current  =  5.5  •*-  &  =  55 
amperes.  By  the  rule  55  X  5.5  X  0.24  =  72.6 
joules  per  second. 

Note. — The  energy  of  a  current  is  given  by  Rules  30,  31 
and  32  in  watts,  so  that  all  cases  are  provided  for  by  a  com- 
bination  of  one  or  the  other  of  these  rules  with  Rule  34 
An  example  will  be  given  for  each  case. 


56  ARITHMETIC  OF  ELECTRICITY. 

A  current  of  .8  ampere  is  sent  through  50  lamps 
in  series,  each  of  137^  ohms  resistance.  What 
heat  does  it  develop  per  second? 

Answer:  The  resistance  =  137^  X  50.  By  rules 
30  and  34  we  have,  rate  of  heat  produced  =  ,82  X 
137^  X  50  =  4400  watts.  4400  X  0.24  =  1056 
joules  per  second. 

Kules  31  and  34.  Fifty  incandescent  lamps,  110 
volt,  137^  ohms,  each  are  placed  in  parallel. 
What  heat  per  second  do  they  develop? 

Solution:  By  Ohm's  law  total  resistance 
•t-  50  =  2.75  ohms.  By  rules  31  and  34  rate  of  heat 
produced  =  HO2  -*-  2.75  =  4400  watts  and  4400  X 
0.24  =  1056  joules  per  second  as  before. 

Rules  32  and  34.  Through  50  incandeseent  110 
volt  lamps  a  current  of  .8  ampere  is  passed,  the 
lamps  being  in  series.  What  heat  per  second  do 
they  develop? 

Solution:  By  rules  32  and  34  rate  of  heat  -=  110  X 
50  X  .8  =  4400  watts  and  1056  joules  per  second  as 
before. 

These  three  examples  are  purposely  made  to  refer 
to  the  same  set  of  lamps,  to  show  that  rules  30,  31, 
and  32  are  identical.  Each  fits  one  of  the  three 
forms  of  statement  of  data.  They  also  are  designed 
to  bring  out  the  fact  that  the  unit  "watts,"  being 
based  partly  on  amperes,  includes  the  id<*a  of  rate, 
not  of  absolute  quantity.  Hence  watts  "per  sec- 
ond "  is  not  stated,  as  it  would  be  meaningless  ox 


WOBK  AND  ENERGY.  57 

redundant,  while  the  joule,  denoting  an  absolute 
quantity,  has  to  be  stated  "per  second"  to  indicate 
tlie  rate.  There  is  such  a  unit  as  an  "ampere- 
second,"  viz.,  the  "coulomb,"  but  there  is  no  such 
thing  as  an  "ampere  per  second,"  or  if  used  it 
means  the  same  as  an  " ampere  per  hour,"  "ampere 
per  day"  or  "ampere."  The  same  applies  to  watts 
and  to  power  units  such  as  "horse-power." 

SPECIFIC  HEAT. 

The  specific  heat  of  a  substance  is  the  ratio  of  its 
capacity  for  heat  to  that  of  an  equal  quantity  of 
water.  It  almost  invariably  is  referred  to  equal 
weights.  Here  it  will  be  treated  only  in  that  con- 
nection. 

The  coefficient  of  specific  heat  of  any  substance  is 
the  factor  by  which  the  specific  heat  of  water  (=  1  or 
unity)  being  multiplied  the  specific  heat  of  the  sub- 
stance is  produced. 

Rule  35.  The  weight  of  any  substance  corresponding 
to  any  number  of  joules  multiplied  by  its  specific  heat 
gives  the  corresponding  weight  of  water,  and  vice  versa. 

EXAMPLE. 

A  current  of  .75  amperes  is  sent  for  5  minutes 
through  a  column  of  mercury  whose  resistance  was 
0.47  ohm.  The  quantity  of  mercury  was  20.25 
grams,  and  its  specific  heat  0.0332.  Find  the  rise 
of  temperature,  assuming  that  no  heat  escapes  by 
radiation.  (Day.) 


58  ARITHMETIC  OF  ELECTRICITY. 

Solution:  By  Eules  30  and  34,  we  find  rate  of  heat 
=  .752  X  .47  =  .264375  watts;  .264375  X  .24  = 
.06345  joules  per  second.  The  current  is  to  last  for 
300  seconds  .'.  total  joules  =  .06345  X  300  =  19.035 
joules.  This  must  be  divided  by  the  specific  heat 
of  mercury  to  get  the  corresponding  weight  of  mer- 
cury; 19.035  -*•  .0332  =  573  gram  degrees  of  mercury. 
Dividing  this  by  the  weight  of  mercury,  20.25  grams, 
we  have  573  •*•  20.25  =  28°  0. 

Rule  36.  By  radiation  and  convection,  4000  joule 
about  is  lost  by  any  unpolished  substance  in  the  air  for 
each  square  centimeter  of  surface,  and  for  each  degree 
that  it  is  heated  above  the  atmosphere. 

EXAMPLE. 

A  conductor  of  resistance,  8  ohms,  has  a  current 
of  10  amperes  passing  through  it.  It  is  1  centi- 
meter in  circumference,  and  100  meters  long.  How 
hot  will  it  get  in  the  air? 

Solution:  By  Eule  30,  etc.,  the  heat  developed  per 
second  in  joules  is  102  X  8  -*-  4.16  =  192.3  joules.  The 
surface  of  the  conductor  in  centimeters  is  10,000  X 
1  =  10,000  sq.  cent.  It  therefore  develops  heat  at 
the  rate  of  192.3  X  1Q-*  =  .01923  joules  per  second 
per  square  centimeter  of  surface.  When  the  loss  by 
radiation  and  convection  is  equal  to  this,  it  will  re- 
main constant  in  temperature.  Therefore  .01923 
•*•  4<jW  =  76.92,  the  number  of  degrees  C.  above  the 
air  to  which  the  conductor  could  be  heated  by  such 
a  current. 


WORK  AND  ENERGY.  69 

Results  of  this  character  are  only  approximate, 
as  the  coefficient,  nsW,  is  not  at  all  accurate. 

Rule  37.  The  cube  of  the  diameter  in  centimeters  of 
a  wire  of  any  giveii  matt-rial  that  will  attain  a  given 
temperature  centigrade  under  a  given  current  is  equal 
to  the  product  of  the  square  of  the  current  in  amperes  x 
Sp.  Resistance  in  microhms  of  the  substance  of  the 
wire,  multiplied  by  .OOO391,  and  divided  by  the  tem- 
perature in  degrees  centigrade. 

s  _  O3  X  Sp.  Bes.  x. 000391 


EXAMPLES. 

A  lead  safety  catch  is  to  be  made  for  a  current  of 
7.2  amperes.  Its  melting  point  is  335°  C.,  and  its 
specific  resistance  19.85  microhms  per  cubic  centi- 
meter. What  should  its  diameter  be?  (Day.) 

Solution:  By  the  rule,  the  cube  of  the  diameter  = 
7.22  X  19.85  X  .00039  -*-  335  =  .001198.  The  cube 
root  of  this  gives  the  diameter  in  centimeters.  It  is 
.10582  or  .106  cm. 

A  copper  wire  is  to  act  as  safety  catch  for  500  am- 
peres: melting  point  1050°  C— Sp.  Kesistance  1.652 
microhms.  What  should  its  diameter  be?  (Day.) 

_,    ,  .,    ,  ,     ,.  500*  x  1.652  x  .000391    _ 

Solution:   Cube  of  diameter  =  -        1050 
!^g§  =  .1523.     The  cube  root  of  this  is  .5341  centi- 

1050 

meter,  the  thickness  of  the  wire  sought  for. 

It  will  be  observed  that  in  this  rule  no  attention 
is  paid  to  the  length  of  the  wire,  as  the  effect  of  a 
current  in  melting  a  wire  has  no  reference  to  its 


60  ARITHMETIC  OF  ELECTRICITY. 

length.     The  time  of  fusion  will  vary  with  the  spe- 
cific heat,  but  will,  of  course,  be  only  momentary, 

WOEK  OF  A  CURRENT. 

Rule  38.  The  work  of  a  current  In  a  given  eircnit  is 
proportional  to  the  volt  amperes.  w  =  EC 

EXAMPLE. 

A  current  A  of  3.5  amperes  with  difference  of 
potential  in  the  circuit  of  20  volts  is  to  be  com- 
pared to  B,  a  3  ampere  current  with  a  difference  of 
potential  of  1000  volts  in  the  circuit;  what  is  the 
ratio  of  work  produced  in  a  unit  of  time? 

Solution:  Work  of  A  :  work  of  B  ::  3.5  X  20  : 
3  X  1000  or  as  70  :  3000  or  as  I  :  42-1fW 

Rule  39.  The  work  of  a  current  in  a  given  circuit  Is 
equal  to  the  volt -coulombs  divided  by  the  acceleration 
of  gravitation  (9.81  meters).  This  gives  the  result  in 
kilogram-meters.  (7.23  foot  pounds.)  -,,_E.  C.t 

~~9^r 

EXAMPLE. 

A  current  of  20  amperes  is  maintained  in  a  circuit 
by  an  E.  M.  F.  of  20  volts.  What  work  does  it  do 
in  one  minute  and  a  half  (90  sec.)? 

Solution :  Work  =  20  X  20  X  90  sec  -4-  9.81  =  3670 
kgmts.  and  3670  X  7.23  =  26,534  foot  pounds. 

Note. — This  is  easily  reduced  to  horse-power.  26,534  foot 
pounds  in  90  sec.  =  17,688  foot  pounds  in  1  min.  1  H.  P. 
—  33,000  foot  pounds  per  min.  .  •.  ||f  £  £  =  .536  H.  P.  of 
above  current  and  circuit.  The  same  result  can  be  ob- 
tained by  Rule  41  thus:  ^  =  .536  H.  P. 


WORK  AND  ENERGY.  61 

Rale  4O.  To  determine  work  done  by  a  current  in  a 
given  circuit  apply  Rules  30,  31  or  32  as  the  case  re- 
quires. These  give  directly  the  watts.  Multiply  by  sec- 
onds and  divide  by  9.81.  The  result  is  kilogram-me- 
ters. 

EXAMPLES. 

10  amperes  are  maintained  for  55  sec.  through  15 
ohms.  What  is  the  work  done? 

Solution  :  By  rule  32,  watts  =  10'  X  15  =  1500. 
Work  =  1500  X  55  -*•  9.81  =  8409  kgmts. 

1000  volts  are  maintained  between  terminals  of  a 
lead  of  20  ohms  resistance.  Calculate  the  work 
done  per  hour. 

Solution:  By  Eule  32  watts  =  10002  -*-  20  =  50,000. 
One  hour  =  3600  sec.  Work  =  50,000  X  3600  •*-  9.81 
=  18,348,623  kgmts. 

These  rules  give  the  basis  for  determining  the  ex- 
pense of  maintaining  a  current.  The  expense  of 
maintaining  a  horse-power  or  other  unit  of  power  or 
work  being  known  the  cost  of  electric  power  is  at 
once  calculable. 

ELECTRICAL  HORSE-POWER. 

Power  is  the  rate  of  doing  work  or  of  expending 
energy.  In  an  electric  circuit  one  horse-power  is 
equal  to  such  a  product  of  the  current  in  amperes, 
by  the  E.  M.  F.  in  volts  as  will  be  equal  to  746. 

Rule  41.  The  electric  horse  power  is  found  by  multi- 
plying  the  total  amperes  of  current  by  the  volts  or  E.  OT. 
F»  of  a  circuit  or  given  part  of  one  and  dividing  by  746. 

LJ     B      —     EC 

H.P.  — 


62  ARITHMETIC  OF  ELECTEICITT. 

EXAMPLE. 

250  incandescent  lamps  are  in  parallel  or  on  mul- 
tiple arc  circuit.  Each  one  is  rated  at  110  volts 
and  220  ohms.  What  electric  H.  P.  is  expended  on 
their  lighting? 

Solution:  The  resistance  of  all  the  lamps  in  par- 
allel is  equal  to  ff&  ohm.  The  current  is  equal  to 
110  +  m  =  125  amperes.  H.  P.  =  125  X  110  -  746 
=  18&  H.  P.  or  13&  lamps  to  the  electrical  H.  P. 

As  it  is  a  matter  of  indifference  as  regards  absorp- 
tion of  energy  how  the  lamps  are  arranged,  a  simpler 
rule  is  the  following,  where  horse-power  required  for 
a  number  of  lamps  or  other  identical  appliances  is 
required. 

Rule  42.  multiply  together  the  voltage  and  amperage 
of  a  single  lamp  or  appliance ;  multiply  the  product  by 
the  11  umber  of  lamps  or  appliances  and  divide  by  746* 

EXAMPLE. 

Take  data  of  last  problem  and  solve  it. 

Solution:  Current  of  a  single  lamp  =  %U  =  %  am- 
pere. H.  P.  =  110  X  ^  X  250  •*•  746  =  ISA  H.  P. 

"When  the  voltage  and  amperage  are  not  given 
directly,  the  missing  one  can  always  be  calculated 
by  Ohm's  law  and  the  above  rules  can  then  be  ap- 
plied. The  same  can  be  done  by  applying  following: 

Rnle  43.  To  determine  the  electrical  horse-power 
apply  Rules  30,  3 1,  or  32;  these  give  directly  the  watts; 
multiplying  the  result  by — -or  dividing  by  746  gives 
i lie  horse-power. 


WORK  AND  ENERGY.  63 

EXAMPLES. 

A  current  of  10  amperes  is  maintained  through  50 
ohms  resistance.  What  is  the  electrical  horse- 
power? 

Solution:  By  rules  30  and  43  we  have  watts  = 
10a  X  50  =  5000  and  electrical  horse-power  =  5000 
-f-  746  or  6.7  H.  P. 

An  electromotive  force  of  1500  volts  is  maintained 
in  a  circuit  of  200  ohms  resistance.  What  is  the 
electrical  horse-power? 

Solution:  By  Rules  31  and  43  we  have  watts  = 
15002  -*-  200  =  11,250.  Electrical  horse-power  = 
11,250  •*•  746  or  15  H.  P.  (nearly). 

Thus  the  volt-amperes  or  watts  are  units  of  rate 
of  heat  energy  or  of  rate  of  mechanical  energy. 
The  ratio  of  joules  per  second  to  a  horse-power  is 
746:  4.16  or  179.3  joules  per  second  =  1  H.  P. 
Other  ratios  of  power  and  heat  units  will  be  found 
in  the  tables. 

DUTY  AND  EFFICIENCY  OF  ELECTRIC  GENERATORS. 

Rule  44.  The  duty  of  an  electric  generator  is  the  quo- 
tient obtained  by  dividing  the  total  electric  energy  by 
the  mechanical  energy  expended  in  turning  the  arma- 
ture. 

-      e.  H.  P.  (total) 
m.  H.  P. 

EXAMPLES. 

A  dynamo  is  driven  by  the  expenditure  of  58 
H.  P.  Its  internal  resistance  is  10.7  ohm.  Th* 


64  ARITHMETIC  OF  ELECTRICITY. 

resistance  of  the  outer  circuit  is  150  ohms  and  it 
maintains  a  current  of  16  amperes.  What  is  its 
duty? 

Solution:  The  total  electrical  H.  P.  is  found  by 
Eules  30  and  43  to  be  162  X  160.7  •*•  746  =  55.1 
H.  P.  Duty  =  55.1-  58.0  =  951 

The  result  must  always  be  less  than  unity;  if  it 
exceeded  unity  it  would  prove  that  there  had  been 
an  error  in  some  of  the  determinations. 

Rule  45.  The  commercial  efficiency  of  a  generator  is 
the  quotient  obtained  by  dividing  the  electric  energy  in 
the  outer  circuit  by  the  mechanical  energy  expended  in 
turning  the  armature. 

_  —-I      e.  H.  P.  (outer  circuit) 
m.H.P. 

EXAMPLES. 

What  is  the  commercial  efficiency  of  the  dynamo 
just  cited? 

Solution:  The  electrical  H.  P.  of  the  outer  circuit 
is  found  by  the  same  rules  to  be  162  X  150  -5-746  = 
51.5  commercial  efficiency  =  51.5  •*•  58.0  =  88.81 

Rule  46.  The  resistance  of  the  outer  circuit  is  to  the 
total  resistance,  as  the  commercial  efficiency  Is  to  the 
duty. 

EXAMPLES. 

Take  the  case  of  the  generator  last  given  and 
from  its  duty  calculate  the  commercial  efficiency. 

Solution:  150:  160.7 ::  x\  95.0.-.  x  =  88.8  or 
88.81 


CHAPTER  VIIL 

BATTERIES. 

GENERAL  CALCULATIONS  OF  CURRENT. 

A  BATTERY  is  rated  by  the  resistance  and  electro- 
motive force  of  a  single  cell,  which  factors  are 
termed  the  cell  constants.  In  the  case  of  storage 
batteries,  whose  susceptibility  to  polarization  is  very 
slight,  the  resistance  is  often  assumed  to  be  neglible. 
It  is  not  so,  and  in  practice  is  always  knowingly  or 
otherwise  allowed  for. 

From  the  cell  constants  its  energy-constant  may 
be  calculated  by  Rule  31,  as  equal  to  the  square  of 
its  electro-motive  force  divided  by  its  resistance. 
This  expresses  its  energy  in  watts  through  a  circuit 
of  no  resistance. 

There  are  two  resistances  ordinarily  to  be  consid- 
ered, the  resistance  of  the  battery  which  is  desig- 
nated by  R  or  by  n  R  if  the  number  of  cells  is  to  be 
implied  and  the  resistance  of  the  external  circuit 
which  is  designated  by  r. 

Rule  47.  The  current  given  by  a  battery  is  equal  to 
its  electro-motive  force  divided  by  the  sum  of  the  exter- 
nal and  internal  resistances. 

C=BT? 


66  ARITHMETIC  OF  ELECTRICITY. 


Six  cells  in  parallel. 


Six  cells  in  series. 


Six  cells— two  in  parallel,  three  in  series. 


Six  cells— three  in  parallel,  two  in  series. 


ARRANGEMENT  OF  BATTERY  CELLS. 


BATTERIES.  67 

EXAMPLE. 

A  battery  of  50  cells  arranged  to  give  75  volts 
E.  M.  F.  with  an  internal  resistance  of  100  ohms 
sends  a  current  through  a  conductor  of  122  ohms 
resistance.  What  is  the  strength  of  the  current? 

Solution:  Current  =  75  -*-  (100  +  122)  =  .338 
ampere.  This  rule  has  already  been  alluded  to 
under  Ohm's  law  (page  14). 

ARRANGEMENT  OF  CELLS  IN  BATTERY. 

In  practice  the  cells  of  a  battery  are  arranged  in 
one  of  three  ways,  a:  All  may  be  in  series;  b:  all 
may  be  in  parallel;  c:  some  may  be  in  series  and  some 
in  parallel,  so  as  to  represent  a  rectangle,  s  cells  in 
series  by  p  cells  in  parallel,  the  total  number  of  cells 
being  equal  to  the  product  of  s  and  p. 

Other  arrangements  are  possible.  Thus  the  cells 
may  represent  a  triangle,  beginning  with  one  cell, 
followed  by  two  in  parallel  and  these  by  three  in 
parallel  and  so  on.  This  and  similar  types  of  arrange- 
ment are  very  unusual  and  little  or  nothing  is  to  be 
gained  by  them. 

Rule  48.  The  electromotive  force  of  a  battery  is 
equal  to  the  E.  .11.  F.  of  a  single  cell  multiplied  by  the 
number  of  cells  in  series. 

Rule  49.  The  resistance  of  a  battery  is  equal  to  the 
number  of  its  cells  in  series,  multiplied  by  the  resist-, 
ance  of  a  single  cell  and  divided  by  the  number  of  its 
cells  in  parallel. 

-,    battery      s  R 

~~ 


68  ARITHMETIC  OF  ELECTRICITY. 

EXAMPLES. 

A  battery  of  50  gravity  cells  1  volt,  3  ohms  each 
is  arranged  10  in  parallel  and  5  in  series.  What  is 
its  resistance  and  electromotive  force? 

Solution:  Resistance  =  5  X  3  -*-  10=  1.5  ohms. 
E.  M.  F.  =  5  X  1  =  5  volts. 

The  same  battery  is  arranged  all  in  parallel;  what 
is  its  resistance  and  E.  M.  F.  ? 

Solution:  This  gives  one  cell  in  series. 

Resistance  =  1  X  3  -s-  50  =  .06  ohms. 
E.  M.  F.     =1X1  =  1  volt. 

The  same  battery  is  arranged  all  in  series;  what  is 
its  resistance? 

Solution:  This  gives  one  cell  in  parallel. 

Resistance  =  5°x3  =  150  ohms. 

E.  M.  F.     =  501  X  1  =  50  volts. 

The  current  given  by  a  battery  is  obtained  from 
these  rules  and  from  Ohm's  law. 

EXAMPLE. 

150  cells  of  a  battery  (cell  constants  1.9  volts, 
$  ohm)  are  arranged  10  in  series  and  15  in  parallel. 
They  are  connected  to  a  circuit  of  1.7  ohms  resist- 
ance. What  is  the  current? 

Solution:  The  resistance  of  the  battery  =  ^p  = 
.333  ohms.  The  E.  M.  F.  =  10  X  1.9  =  19  volts. 
Current  =  19-5-  (.333  +  1.7)  =  9.34  amperes. 


BATTERIES.  69 

CELLS  REQUIRED  FOR  A  GIVEN  CURRENT. 

To  calculate  the  cells  required  to  produce  a  given 
current  through  a  given  resistance  and  the  arrange- 
ment of  the  cells  proceed  as  follows. 

Rule  50.  Calculate  the  cell  current  through  zero  exter- 
nal resistance.  Case  A.  If  it  is  twice  as  great  or  more 
than  twice  as  great  as  the  current  required  apply  .Rule 

51.  Case  I*.   If  less  than  twice  as  great  and  more  than 
equal  or  less  than   equal  and  more   than  one  half  as 
great  as  the  current    required  and    so  on  apply  Rule 

52.  Case  C.    If  the  cell  current  is  equal  to  or  is  a  unit* 
ary  fraction  (i,  i,  1,  etc.)  of  the  current  required  apply 
Rule  53.  458 

Rule  51.  Case  A.  Divide  the  required  difference  of  po- 
tential of  the  outer  circuit  by  the  voltage  of  a  single  cell 
diminished  by  the  product  of  the  required  current  mul- 
tiplied by  the  resistance  of  a  single  cell*  Arrange  the 
cells  in  series. 


EXAMPLES. 

Five  lamps  in  parallel,  each  of  100  volts  200  ohms, 
are  to  be  supplied  by  a  battery  whose  cell  constants 
are  2  volts  i  ohm.  How  many  cells  and  what 
arrangement  are  required? 

Solution:  Cell  current  =  |  =  10  amperes.  The 
resistance  of  the  five  lamps  in  parallel  (Rule  12)  = 
2£2  =  4Q  0}lms>  The  required  current  therefore  =• 
W  =  21  amperes.  As  10  exceeds  21  X  2  (Rule  50)  it 
falls  under  case  A.  By  Rule  51  the  number  of  cells 
is  8_*ftXt)  =  ^  =  66.6  or  67  cells,  as  a  cell  cannot 
be  divided.  The  cells  must  be  in  series. 


70  ARITHMETIC  OF  ELECTRICITY. 

Proof:  The  E.  M.  F.  of  the  67  cells  in  series 
=  67  X  2  =  134  volts;  their  resistance  =  67  X  £  = 
13.4  ohms.  The  resistance  of  the  lamps  in  par- 
allel is  40  ohms.  Hence  by  Ohm's  law  the  cur- 
rent =  ^  + 13  4-  =  2.51  amperes,  the  current  required. 

The  same  lamps  are  placed  in  series.  Calculate 
the  cells  of  the  same  battery  required.  Cell  current 
=  10  amperes.  Current  required  =  ^Xg  or  |  am- 
pere. As  10  exceeds  *  X  2  (Rule  50)  it  falls  again 
under  case  A.  By  Eule  51  cells  required  =  a_^xi) 
=  263. 

Proof:  Current  =  -52^1(m  =  I  ampere  the  current 
required.  526  is  the  number  of  cells  multiplied  by 
the  voltage  of  one  cell;  52.6  is  the  number  of  cells 
multiplied  by  the  resistance  in  ohms  of  a  single  cell; 
1000  is  the  resistance  of  a  single  lamp,  200  ohms., 
multiplied  by  the  number,  5,  of  lamps  in  series. 

"Whenever  the  arrangement  and  number  of  cells  of 
a  battery  has  been  calculated  the  calculation  should 
be  proved  as  above. 

Rnle  52.  Case  15.  Group  two  or  more  cells  In  parallel 
so  as  to  obtain  by  calculation  from  them  through  no  ex- 
ternal resistance  a  current  twice  as  great  or  more  than 
twice  as  great  as  the  required  current.  Then  treating 
the  group  as  if  it  was  a  single  cell  apply  Rule  51  to  de- 
termine the  number  of  groups  in  series. 

EXAMPLE. 
Assume  the  same  lamps  in  parallel,  requiring  the 


BATTEEIES.  71 

current  already  calculated  of  2i  amperes.  Assume 
a  battery  of  constants  1  volt  .25  ohm,  giving  a 
cell  current  of  4  amperes.  This  is  less  than  2$  X  2 
and  more  than  2i  X  l;  therefore  it  falls  under  Case 
B. 

Solution:  A  group  of  two  cells  in  parallel  gives 
.T^S  =  8  amperes.  8  exceeds  2i  X  2  . '.  applying  .Rule 
49  we  have  number  of  groups  =  100  -*-[!—  (2i  X 
.125)]  =  146  groups  in  series.  Total  number  of  cells 
=  2  in  parallel,  146  in  series  =  292  cells. 

Proof:  Current  =  146  -H  (40  +  18.25)  =2.5  am- 
peres. 

Rule  53.  Case  C.  Place  as  many  cells  in  series  as  will 
give  twice  the  required  voltage.  Place  as  many  cells  in 
parallel  as  will  give  a  resistance  equal  to  that  of  the 
external  circuit. 

EXAMPLE. 

Assume  the  same  lamps  in  parallel.  Assume  a 
battery  of  cell  constants,  1  volt,  4  ohms.  The  lamp 
current  is  21  amperes.  The  cell  current  is  *  ampere. 
The  cell  current  therefore  equals  (±  -*•  2|)  &  of  the 
required  current.  This  falls  under  Case  C.  and  is 
solved  by  Eule  53. 

Solution:  Voltage  required  100.  By  the  rule 
cells  in  series  =  100  X  2  ==  200.  These  have  a  re- 
sistance of  800  ohms.  To  reduce  this  to  the  resist- 
ance of  the  outer  circuit,  viz.,  40  ohms,  800  -f-  40 
=  20  cells  must  be  placed  in  parallel.  Total  cells  =- 
20  x  200  =  4000  cells. 


72  ARITHMETIC  OF  ELECTRICITY. 

Proof:  Current  =  200  •*-  (40  +  40)  =  2.5. 

Rule  54.  All  cases  coming  under  Case  C.  may  be  sim- 
ply solved  lor  tlie  total  number  of  cells  by  dividing  tli« 
external  energy  by  tlie  cell  energy  and  multiplying  by  4. 
'I'll is  gives  tiic  number  of  cells. 

EXAMPLE. 

Take  as  cell  constants  .75  volt  M  ohm  giving  i  am- 
pere. Assume  20  lamps,  each  50  volts,  50  ohms  and 
1  ampere.  As  i  -*-  1  is  a  unitary  fraction  (£)  Case 
C.  applies. 

Solution:  Cell  energy  =  i  X  .75  =  .375  watts.  Ex- 
ternal energy  =  50  x  1  X  20  =  1000  watts.  (1000  •*- 
.375)X  4-10,666  cells. 

Solution  by  Rule  53:  Voltage  required  taking 
lamps  in  series  —  20  x  50  =  1000.  To  give  twice 
this  voltage  requires  2000  •*-  .75  =  2667  cells  in 
series  whose  resistance  is  2667  X  1.5  =  4000  ohms. 
To  reduce  this  to  1000  ohms  we  need  4  such  series 
of  cells  in  parallel  giving  10,668  cells. 

Proof:  Current  =  ^£  +  1000  =  1  ampere. 

Slight  discrepancies  will  be  noticed  in  the  current 
strength  given  by  different  calculations.  This  is 
unavoidable  as  a  cell  cannot  be  fractioned  or  di- 
vided. 

EFFICIENCY  OF  BATTERIES. 

Rule  55.  The  efficiency  of  a  battery  is  expressed  by  di- 
viding the  resistance  of  the  external  circuit  by  the  total 
resistance  of  the  circuit. 

Efficiency  =  ^- . — 
xv  -t-  r 


BATTERIES.  73 

EXAMPLE. 

A  battery  consists  of  67  cells  in  series  of  constants 
2  volts  i  ohm.  It  supplies  5  lamps  in  parallel,  each 
100  volts  200  ohms  constants.  What  is  its  effi- 
ciency? 

Solution:  The  resistance  of  the  battery  is  67  X  | 
=  13.4  ohms.  The  resistance  of  the  lamps  is  (Rule 
12)  *$*  =  40  ohms.  Therefore  the  efficiency  of  the 
battery  is  40.0  -*-  (40  +  13.4)  =  .749  or  74.9*. 

Rule  5  6.  To  calculate  the  number  of  battery  cells  and 
their  arrangement  for  a  given  efficiency :  Express  the 
efficiency  as  a  decimal,  multiply  the  resistance  of  the 
external  circuit  by  the  complement  of  the  efficiency 
(1— efficiency)  and  divide  the  product  by  the  efficiency; 
this  gives  the  resistance  of  the  battery.  Add  the  two 
resistances  and  multiply  their  sum  by  the  current  to  be 
maintained  for  the  K.  M.  F.  of  the  battery.  Arrange  the 
cells  accordingly  as  near  as  possible  to  these  require- 
ments. 

EXAMPLES. 

Five  lamps,  each  100  volts  200  ohms  in  parallel 
are  to  be  supplied  by  a  battery  of  cell  constants  2 
volts  .4  ohm.  The  efficiency  of  the  battery  is  to  be 
as  nearly  as  possible  75£.  Calculate  the  number  of 
cells  and  their  arrangement. 

Solution:  The  constants  of  the  external  circuit  are 
40  ohms  (Rule  12)  and  100  volts.  Applying  the 
rule  we  have  [40  X  (1— .75)]  •*-  .75  =  ^  =  13* 
ohms,  the  resistance  of  the  battery.  By  Ohm's  law 
the  E.  M.  F.  of  the  battery  =  (40  +  13*)  X  2.5  = 


74  ARITHMETIC  OF  ELECTRICITY. 

133*   volts.     These  constants,    13*   ohms  and  133* 
volts,  require  67  cells  in  series  and  2  in  parallel. 

Proof:  a.  Of  efficiency,  by  Rule  55,  4Q^13i  =  .75 
or  75#.  b.  Of  number  of  cells  and  of  their  arrange- 
ment 67  X  2  =  134  volts;  (67  X  .4)  -*-  2  =  13.4 
ohms;  134  •*-  (13.4  +  40)  =  2.5  amperes. 

Rule  57.  Where  a  fractional  or  mixed  number  of  cells 
in  parallel  are  called  for  to  produce  a  given  efficiency, 
take  a  group  of  the  next  highest  integral  number  of 
cells  in  parallel  and  proceed  as  in  Rule  51. 

EXAMPLE. 

Assume  a  current  of  3*  amperes  to  be  supplied 
through  a  resistance  of  30  ohms,  absorbing  100 
volts  E.  M.  F.  Let  the  cell  constants  of  a  battery 
to  supply  this  circuit  be  2  volts,  *  ohm.  Calculate 
the  cells  and  their  arrangement  for  80  per  cent, 
efficiency. 

Solution:  By  Rule  56  efficiency  =  .80  and 
80  X  so"'80*  =  ^  ohms,  which  is  the  required  resist- 
ance of  the  battery;  7*+  30  =  37*  ohms  are  the  total 
resistance  of  the  circuit.  By  Ohm's  law,  37*  X  3*  = 
125  volts,  the  required  E.  M.  F.  of  the  battery. 
This  requires  63  cells  in  series,  with  a  resistance  for 
one  series  of  63  X  *  =  10*  ohms.  To  reduce  this  to 
7^  ohms  ^  =  1.4  cells  in  parallel  are  required.  As 
this  is  a  mixed  number  we  take  the  next  highest  in- 
tegral number  and  place  2  cells  in  parallel.  The 
constants  of  this  group  of  2  cells  are  2  volts,  & 


BATTERIES.  75 

ohm.  Applying  Rule  51  we  have  for  the  number  of 
such  groups  in  series;  2_(™x  A)  =  58  groups  in  series. 
As  there  are  2  cells  in  parallefthe  total  cells  are  116, 
of  resistance,  58  X  A  =  4.83  ohms,  and  of  E.  M.  F., 
58  X  2  =  116  volts. 

Proof:  Of  efficiency  by  Rule  55,  ^TtM  =  86.1*. 
Of  number  and  arrangement  of  cells  M"^  =  3.33 
amperes. 

It  is  to  be  observed  that  the  efficiency  thus 
obtained  is  far  from  what  is  required.  In  most 
cases  accuracy  can  only  be  attained  by  arranging 
the  battery  irregularly,  which  is  unusual  in  prac- 
tice. An  example  will  be  found  in  a  later  chapter. 

CHEMISTRY  OF  BATTERIES. 

One  coulomb  of  electricity  will  set  free  .010384 
milligrams  of  hydrogen.  The  corresponding  weights 
of  other  elements  or  compounds  are  found  by  multi- 
plying this  factor  by  the  chemical  equivalent,  and 
dividing  by  the  valency  of  the  element  or  metal  of 
the  base  of  the  compound  in  question. 

An  element  or  other  substance  in  entering  into 
any  chemical  combination  develops  more  or  less 
heat,  always  the  same  for  the  same  weight  and  com- 
bination. The  atomic  weight  of  an  element  or  the 
molecular  weight  of  a  compound  divided  by  the  val- 
ency of  the  element  or  metal  of  its  base  gives  the 
original  chemical  equivalent. 


76  ARITHMETIC  OF  ELECTRICITY. 

The  quantities  of  heat  evolved  by  the  combination 
of  quantities  of  substances  expressed  by  their  original 
chemical  equivalents  multiplied  by  one  gram  are 
termed  the  thermo-electric  equivalents  of  the  ele- 
ments or  substances  in  question.  In  the  tables  it  is 
expressed  in  kilogram  degrees  0.  of  water  (kilogram- 
calories). 

From  the  thermo-electric  equivalent  of  a  combi- 
nation we  find  the  volts  evolved  by  it  or  absorbed  by 
the  reciprocal  action  of  decomposition. 

Rule  58.  The  volts  evolved  by  any  chemical  com- 
bination or  required  for  any  chemical  decomposition 
are  equal  to  the  thermo-electric  equivalent  in  kilo* 
gram-calories  multiplied  by  .043. 

E  =  .043  X  H. 

EXAMPLES. 

What  number  of  volts  is  required  to  decompose 
water  ? 

Solution:  From  the  table  we  find  that  the  com- 
bination of  one  gram  of  hydrogen  with  eight  grams 
of  oxygen  liberates  34.5  calories.  Then  34.5  X 
.043  =  1.48  volts. 

Rule  59.  To  determine  the  voltage  of  a  galvanic 
couple  subtract  the  kilogram  calories  corresponding  to 
decompositions  in  the  cell  from  those  corresponding  to 
combinations  in  the  cell  for  effective  energy  and  multi- 
ply by  .043  for  volts. 

EXAMPLES. 

Calculate  the  voltage  of  the  Smee  couple. 
Solution:  In   this  battery     zinc   combines    with 


BATTERIES.  77 

oxygen,  giving  out  43.2  calories  and  combines  with 
sulphuric  acid,  giving  11.7  more  calories;  a  total  of 
54.9  calories.  An  equivalent  amount  of  water  is  at 
the  same  time  decomposed  acting  as  counter-energy 
of  34.5  calories.  The  effective  energy  is  54.9  — 
34.5  =  20.4  calories.  The  voltage  =  20.4  X  .043  = 
.877  volts. 

Calculate  the  voltage  of  the  sulphate  of  copper 
battery. 

Solution:  Here  we  have  combination  of  zinc  with 
sulphuric  acid  as  above  54.9  calories;  decomposition 
of  copper  sulphate  19.2  +  9.2  =  28.4  .-.  54.9  —  28.4 
=  26.5  calories  effective  energy  26.5  X  .034  =  1.1^ 
volts. 

It  will  be  noticed  that  these  results  are  approxi- 
mate. Some  combinations  are  omitted  in  them  as 
either  of  unknown  energy,  or  of  little  importance. 

WORK  OF  BATTERIES. 

The  rate  of  work  of  a  battery  is  proportional  to 
the  current  multiplied  by  the  electro-motive  force. 
The  work  is  distributed  between  the  battery  and  the 
external  circuit  in  the  ratio  of  their  resistances  as  by 
Eule  55.  The  horse-power,  and  heating  power  are 
calculated  by  Rules  30-43,  care  being  taken  to  dis- 
tribute the  energy  acording  to  the  resistance  by 
the  following  rule: 

Rule  6O.    The  effective   rate  of  work   or  the   rate  of 
work  in  the  external  circnit  of  a  battery,  is  equal  to  t  lie 


78  ARITHMETIC  OF  ELECTRICITY. 

total  rate  multiplied  by  the  efficiency  of  the  battery  ex- 
pressed decimally. 

EXAMPLE. 

25  cells  of  2  volt  1  ohm  battery  are  arranged  in 
series  on  an  external  circuit  of  250  ohms  resistance. 
What  work  do  they  do  in  that  circuit? 

Solution:  The  current  (Ohm's  law)  =  ~^|  =  1 


1.818  amperes.  Total  rate  of  work  =  1.818  X  50 
volts  =  90.9  watts.  Efficiency  of  battery  =  m  =  90 
per  cent,  (nearly).  Effective  rate  of  work  =  1.818  X 
50  X  .90  =  81.81  watts. 

CHEMICALS  CONSUMED  IN  A  BATTERY. 

Rule  61.  The  chemicals  consumed  in  grams  by  a  bat- 
tery for  one  kilogram-meter  (7.23  foot  Ibs.)  of  work  are 
found  by  multiplying  the  combining  equivalent  of  the 
chemical  by  the  number  of  equivalents  in  the  reaction 
by  the  constant  .OOO1O1867  and  dividing  by  the  pro- 
duct of  the  10.  M.  F.  by  the  valency  of  the  element  In 
question. 

—  -  _  Equiv.  X  n  X  .000101867 
E  x  valency 

EXAMPLES. 

What  is  the  consumption  of  zinc  and  sulphate  of 
copper  per  kilogram-meter  of  work  in  a  Daniel's 
battery? 

Solution:  Take  the  E.  M.  F.  as  1.07  volt.  The 
equivalent  of  zinc,  a  dyad,  is  65  and  one  atom  en- 
ters into  the  reaction.  The  zinc  consumed  there- 

_  65  XIX.  0—  = 


BATTERIES.  79 

The  equivalent  of  copper  sulphate,  is  159.4.  One 
equivalent  enters  into  the  reaction  carrying  with  it 
one  atom  of  the  dyad  metal  copper.  The  weight 
consumed  therefore  =  m"  *^xxf1M867  =  .0076 
grams.  Add  56.46$  for  water  of  crystallization. 

.  All  these  quantities  are  for  one  kilogram-meter  of 
work  (7.23  foot  Ibs.)  which  may  be  more  or  less 
effective  according  to  circumstances  as  developed  in 
Rules  44,  45,  and  60. 

DECOMPOSITION  OF  COMPOUNDS  BY  THE  BATTERY. 

In  cases  where  a  compound  has  to  be  decomposed 
by  a  battery  two  resistances  may  be  opposed  to  the 
work.  One  is  the  ohmic  resistance  of  the  solution, 
which  is  calculated  by  Rule  16.  The  other  is  the 
electromotive  force  required  to  decompose  the  solu- 
tion. This  is  best  treated  as  a  counter-electromotive 
force.  Then  from  the  known  data  the  current  rate 
is  calculated,  and  from  the  electro-chemical  equiva- 
lents the  quantity  of  any  element  deposited  by  a 
given  number  of  coulombs  is  determined. 

Rule  62.  To  calcnlate  the  metal  or  other  element  lib- 
erated by  a  given  current  per  given  time  proceed  as  fol- 
lows: Calcnlate  the  resistance.  Determine  the  counter- 
electromotive  force  of  the  solution  by  Rule  58  and  sub- 
tract it  from  the  E.  Iff.  F.  of  the  battery  or  generator. 
Apply  Ohm's  law  to  the  effective  voltage  thus  deter- 
mined and  to  the  calculated  resistances  to  find  the  cur- 
rent, multiply  the  electro-chemical  equivalent  of  the 
element  by  the  coulombs  or  ampere-seconds. 


80  ARITHMETIC  OF  ELECTRICITY. 

EXAMPLE. 

A  bath  of  sulphate  of  copper  is  of  specific  resis- 
tance 4  ohms.  The  electrodes  are  supposed  to  be 
10,000  sq.  centimeters  in  area  and  5  centimeters 
apart.  Two  large  Bunsen  elements  in  series  of 
1.9  volts  .12  ohms  each  are  used.  What  weight 
in  milligrams  of  copper  will  be  deposited  per 
hour? 

Solution:  By  Rule  16  the  resistance  of  the  solu- 
tion is  ^j|  =  0.023.  The  electro-chemical  equiva- 
lent of  copper  is  .00033  grams.  The  thermo-electric 
equivalent  for  copper  from  sulphate  of  copper 
is  19.2  -f  9.2  =  28.4  calories.  The  E.  M.  F. 
corresponding  thereto  =  28.4  X  .043  =  1.22  volts 
counter  E.  M.  F.  The  E.  M.  F.  of  the  bat- 
tery =  1.9  X  2  =  3.8  volts,  giving  an  effective  E.  M. 
F.  of  3.8  —  1.22  =  2.58  volts.  The  resistance  of  the 
battery  =  .12  X  2  =  .24  ohms.  The  current  = 
=  9.8  amperes.  This  gives  per  hour  9.8  X 


3,600  =  35,280  coulombs,  and  for  copper  deposited 
.00033  X  35,280  =  11.64  grams. 

In  many  cases  one  electrode  is  made  of  the  mater- 
ial to  be  deposited  and  being  connected  to  the  car- 
bon end  of  the  battery  or  generator  is  dissolved  as 
fast  as  the  metal  is  deposited.  In  such  case  there 
is  no  counter  electro-motive  force  to  be  allowed 
for. 


BATTEEIES.  81 

EXAMPLE. 

Take  the  last  case  and  assume  one  electrode  (the 
anode)  to  be  of  copper  and  to  dissolve.  Calculate 
the  deposit. 

Solution:  Current  =  3.8  -*-  (.24  +  .023)  =  14.4 
amperes  =  51,840  coulombs  per  hour  =  .00033  X 
51,840  =  17.10  grams  of  copper. 


CHAPTER  IX. 

ELECTRO-MAGNETS,  DYNAMOS  AND  MOTORS. 

THE  MAGNETIC  FIELD    AND  LINES  OF  FORCE. 

A  CURRENT  of  electricity  radiates  electro-magnetic 
wave  systems,  and  establishes  what  is  known  as  a 
field  of  force.  The  field  is  more  or  less  active  or  in- 
tense according  to  the  force  establishing  it.  The 
intensity  of  a  field  is  for  convenience  expressed  in 
LINES  OF  FORCE.  These  are  the  units  of  magnetic 
intensity,  often  called  units  of  magnetic  flux,  and 
the  line  as  a  unit  is  comparable  to  the  ampere  X  10, 
which  is  the  C.  G-.  S.  unit  of  current.  A  line  of 
force  is  that  quantity  of  magnetic  flux  which  passes 
through  every  square  centimeter  of  normal  cross- 
section  of  a  magnetic  field  of  unit  intensity.  The 
line  is  at  right  angles  to  the  plane  of  normal  cross- 
section  of  such  field.  Such  intensity  of  field  exists 
at  the  center  of  curvature  of  an  arc  of  a  circle  of  ra- 
dius 1  centimeter,  and  whose  length  is  1  centimeter, 
when  a  current  of  10  amperes  passes  through  this 
arc.  Practically  it  is  the  amount  which  passes 
through  an  area  of  one  square  centimeter,  situated 
in  the  center  of  a  circle  10  centimeters  in  diameter, 


ELECTRO-MAGNETS,  DYNAMOS  AND  MOTORS.   83 

surrounded  by  a  wire  through  which  a  current  of 
7.9578  amperes  is  passing.  The  plane  of  the  circle 
is  a  cross-sectional  plane  of  the  field;  a  line  perpen- 
dicular to  such  plane  gives  the  direction  of  the  lines 
of  force,  or  of  the  magnetic  flux. 

This  cross-sectional  area  is  often  spoken  of  as  the 
field  of  force.  As  a  field  exists  wherever  there  are 
lines  of  force,  there  are  in  each  magnetic  circuit 
either  an  infinite  number  of  fields  of  force,  or  a 
field  of  force  is  a  volume  and  not  an  area. 

The  number  of  lines  of  force  or  of  magnetic  flux 
per  unit  cross-sectional  area  of  the  magnetic  cir- 
cuit, i.  e.  per  unit  area  of  magnetic  field,  expresses 
the  intensity  of  the  field.  In  soft  iron,  it  may  run  as 
high  as  20,000  or  more  lines  per  square  centimeter 
of  cross-section  of  the  iron  which  is  magnetized. 

Just  as  we  might  speak  of  a  bar  of  copper  acting 
as  conductor  for  20,000  C.  G.  S.  units  of  current,  or 
2000  amperes,  so  we  may  speak  of  the  iron  core  of  a 
magnet  carrying  20,000  lines  of  force. 

PERMEANCE  AND  RELUCTANCE. 

This  action  of  centralizing  in  its  own  material 
lines  of  force  is  analogous  to  "conductance."  It  is 
termed  PERMEANCE.  Its  reciprocal  is  termed 
RELUCTANCE,  which  is  precisely  analogous  to 
"  resistance. "  Iron,  nickel,  and  cobalt  possess  high 
permeance;  the  permeance  of  air  is  taken  as  unity. 
At  a  low  degree  of  magnetization,  soft  iron  pos- 


84  ARITHMETIC  OF  ELECTRICITY. 

sesses  10,000  times  the  permeance  of  air.  At  high 
degrees  of  magnetization,  it  possesses  much  less  in 
comparison  with  air,  whose  permeance  is  unchanged 
under  all  conditions. 

There  is  no  substance  of  infinitely  high  reluctance, 
which  is  the  same  as  saying  that  there  is  no  insula- 
tor of  magnetism. 

MAGNETIZING  FORCE    AND  THE  MAGNETIC   CIR- 
CUIT. 

The  producing  cause  of  the  magnetic  flux  or  mag- 
netization just  described  is  in  practice  always  a  cur- 
rent circulating  around  an  iron  core.  The  name  of 
MAGNETIZING  FORCE  is  often  given  to  it.  It  is  the 
analogue  of  electro-motive  force,  and  is  measured  by 
the  lines  of  force  it  establishes  in  a  field  of  air  of 
standard  area. 

A  high  value  for  the  magnetic  force  is  585  lines 
per  square  centimeter.  It  is  proportional  to  the 
amperes  of  current  and  to  the  number  of  turns  the 
conductor  makes  around  it.  Its  intensity  is  often 
given  in  ampere-turns. 

Magnetization  always  implies  a  circuit.  As  far  as 
known,  magnetic  lines  of  force  cannot  exist  without 
a  return  circuit,  exactly  like  electric  currents.  But 
owing  to  the  imperfect  reluctance  of  all  materials, 
the  lines  of  force  can  complete  their  circuit  through 
any  substance*.  In  a  bar  magnet  the  return  branch 
of  the  circuit  is  through  air. 


ELECTRO-MAGNETS,  DYNAMOS  AND  MOTORS.   85 

In  the  same  magnetic  circuit,  the  planes  of  nor- 
mal cross-section  lie  at  various  angles  with  each 
other. 

The  law  of  a  magnetic  circuit  is  exactly  compar- 
able to  Ohm's  law.  It  is  as  follows: 

Rale  63.  The  magnetization  expressed  in  lines  of 
force  is  equal  to  the  magnetizing  force  divided  by  the 
reluctance  or  multiplied  by  the  permeance  of  the  entire 
circuit. 

This  rule  would  he  of  very  simple  application,  ex- 
cept for  the  fact  that  reluctance  increases,  or  per- 
meance decreases,  with  the  magnetization,  and  the 
rate  of  variation  is  different  for  different  kinds  of 
iron. 

Rule  64.  Permeability  is  the  ratio  of  magnetization 
to  magnetizing  force,  and  is  obtained  by  dividing  mag- 
netization by  magnetizing  force. 

Permeability  has  to  he  determined  experimentally 
for  each  kind  of  iron.  It  is  simply  the  expression 
of  a  ratio  of  two  systems  of  lines  of  force.  It 
always  exceeds  unity  for  iron,  nickel,  and  cobalt. 
The  specific  susceptibility  of  any  particular  iron  to 
magnetization  is  its  permeability.  The  susceptibil- 
ity of  a  portion,  or  of  the  whole  of  a  magnetic  cir- 
cuit is  its  permeance. 

GENERAL  RULES  FOR  ELECTRO-MAGNETS. 

The  traction  of  a  magnet  is  the  weight  it  can 
sustain  when  attached  to  its  armature.  It  is  pro- 


86  ARITHMETIC  OF  ELECTRICITY. 

portional  to  the  square  of  the  number  of  lines  of 
force  passing  through  the  area  of  contact. 

Rule  65.  The  traction  of  a  magnet  In  pounds  is  equal 
to  the  square  of  the  number  of  lines  of  force  per  square 
inch,  multiplied  by  the  area  of  contact  and  divided  by 
72,134,000.  In  centimeter  measurement  the  traction 
in  pounds  is  equal  to  the  square  of  the  number  of  lines 
of  force  per  square  centimeter  multiplied  by  the  area  of 
contact  and  divided  by  11,183,000.  The  traction  in 
grams  is  equal  to  the  latter  dividend  divided  by  24,655 
«.s  JT  A  981);  for  dynes  of  traction  the  divisor  is  25.132 


EXAMPLES. 

A  bar  of  iron  is  magnetized  to  12,900  lines  per 
square  inch;  its  cross-section  is  3  square  inches. 
What  weight  can  it  sustain,  assuming  the  armature 
not  to  change  the  intensity  of  magnetization? 

Solution:  12,9002  X  3  =  499,230,000.  This  di- 
vided by  72,134,000  gives  6.914  Ibs.  traction. 

A  table  calculated  by  this  rule  is  given.  A 
diminished  area  of  contact  sometimes  increases  trac- 
tion, and  a  non-uniform  distribution  of  lines  may 
occasion  departures  from  it.  The  above  rule  and 
the  table  alluded  to  are  practically  only  accurate 
for  uniform  conditions.  The  reciprocal  of  the  rule 
is  applied  in  determining  the  lines  of  force  of  a 
magnet  experimentally. 

Rule  66.  The  lines  of  force  which  can  pass  through  a 
magnet  core  with  economy  are  determined  by  the  tables, 
keeping  in  mind  that  it  is  not  advisable  to  let  the  per- 
meability fall  below  200—  30O.  From  them  a  number 
is  taken  (40,000  lines  per  square  inch  for  cast  iron  or 


ELECTRO-MAGNETS,  DYNAMOS  AND  MOTORS.   87 

1OO,000  lines  per  square  incli  for  wrought  iron  are  good 
general  averages)  and  is  multiplied  by  the  cross-sec- 
tional area  of  the  magnet  core. 

Rule  67.  To  calculate  the  magnetizing  force  in  am- 
pere turns  required  to  force  a  given  number  of  magnetic 
lines  through  a  given  permeance,  multiply  the  desired 
lines  of  force  by  the  reluctance  determined  as  below. 

Rule  68.  a.  The  reluctance  of  a  core  or  of  any  portion 
thereof  for  inch  measurements  is  equal  to  the  product 
of  the  length  of  the  core  or  of  the  portion  thereof  by 
O.3132  divided  by  the  product  of  its  cross-sectional  area 
and  permeability. 

6.  The  reluctance  for  centimeter  measurements  is 
equal  to  the  length  of  the  core  divided  by  the  product  of 
1.2566,  by  the  cross-sectional  area  and  the  permea- 
bility. 

EXAMPLES. 

440,000  lines  are  to  be  forced  through  a  bar  of 
wrought  iron  10  inches  long  and  4  square  inches  in 
area;  calculate  its  reluctance  and  the  magnetizing 
force  in  ampere  turns  required  to  effect  this  mag- 
netization. 

Solution:  The  reluctance  (a)  =  10  X  .3132  •*-  (4  X 
permeability).  440,000  lines  through  4  square  inches 
area  is  equal  to  110,000  lines  through  1  square 
inch;  for  this  intensity  and  for  wrought  iron  the 
permeability  =  166.  166  X  4  =  664.  The  reluc- 
tance therefore  =  3.132  -*•  664  =  .0047.  The  mag- 
netizing force  in  ampere  turns  =  440,000  X  .0047 
=  2068. 

The  same  number  of  lines  are  to  be  forced 
through  a  bar  25.80  square  centimeters  area  and 


88  ARITHMETIC  OF  ELECTRICITY. 

25.40  centimeters    long.       Calculate    the    ampere 
turns. 

Solution:  440,000  lines  through  25.80  sq.  cent. 
=  17,054  through  1  sq.  cent.,  for  which  the 


permeability  =  161.  The  reluctance  therefore,  (b) 
=  25.40  -  (1.2566  X  25.80  X  161)  =.0048.  The 
ampere  turns  =  440,000  X  .0048  =  2112. 

MAGNETIC  CIRCUIT  CALCULATIONS. 

Practically  useful  calculations  include  always  the 
attributes  of  a  full  magnetic  circuit,  because  mag- 
netization can  no  more  exist  without  a  circuit  than 
can  an  electric  current.  In  practice  an  electro- 
magnetic circuit  consists  of  four  parts:  1,  The 
magnet  cores;  2  and  3,  the  gaps  between  armature 
and  magnet  ends;  4,  the  armature  core.  To  cal- 
culate the  relations  of  magnetizing  force  to  magne- 
tization the  sum  of  the  reluctances  of  these  four 
parts  has  to  be  found.  A  further  complication 
is  introduced  by  leakage.  The  permeability  of  well 
magnetized  iron  being  so  low,  not  exceeding  150  to 
300  times  that  of  air,  a  quantity  of  lines  leak  across 
through  the  air  from  magnet  limb  to  magnet  limb. 
Leakage  is  included  in  the  sum  of  the  reluctances 
by  multiplying  the  reluctance  of  the  magnet  core  by 
the  coefficient  of  leakage,  which  is  calculated  for 
each  case  by  more  or  less  complicated  methods.  For 
parallel  cylindrical  limb  magnets  the  calculation  is 


ELECTRO-MAGNETS,  BYNAMOS  AND  MOTORS.  89 

exceedingly  simple.  The  calculation  in  all  cases  is 
simplified  by  the  fact  already  stated,  that  in  air 
permeability  is  always  equal  to  unity,  whatever  the 
degree  of  magnetization.  For  copper  and  other 
non-magnetizable  metals  the  variation  from  unity  is 
so  slight  that  it  may,  for  practical  calculations,  be 
treated  as  unity. 

LEAKAGE  OF  LIKES  OF  FORCE. 

Leakage  is  the  magnetic  flux  through  air  from 
surfaces  at  unequal  magnetic  potential,  such  as 
north  and  south  poles  of  magnets.  It  is  measured 
by  lines  of  force  and  is  proportional  to  the  relative 
permeance  of  its  path. 

The  coefficient  of  leakage  of  a  magnetic  circuit  is 
the  quotient  obtained  by  dividing  the  total  magnetic 
flux  by  the  flux  through  the  armature.  The  total 
magnetic  flux  is  the  maximum  flux  through  the 
magnet  core. 

Rule  69.  To  obtain  the  coefficient  of  leakage  divide 
the  permeance  of  the  armature  core  and  of  the  two  gaps 
plus  one-half  the  permeance  of  air  between  magnet 
limbs  by  the  permeance  of  the  armature  core  and  of  the 
two  gaps. 

EXAMPLE. 

The  total  flux  through  an  armature  core  is  found 
to  be  at  the  rate  of  70,000  lines  per  square  inch,  and 
the  armature  core  is  3  inches  diameter  and  10  inches 
long.  The  average  length  of  travel  of  the  magnetic 


90  ARITHMETIC  OF  ELECTRICITY. 

lines  through  it  is  H  inches.  The  air  gaps  are  10  X 
3  inches  area  and  i  inch  thick.  The  permeance  be- 
tween the  limbs  of  the  magnet  is  500.  Calculate 
the  coefficient  of  leakage. 

Solution:  70,000  lines  per  square  inch  gives  a 
permeability  of  1,921.  By  Kule  68  the  reluctance 
of  the  armature  core  is  ^  J*im  X  .3132  =  .000008. 
The  reluctance  of  a  single  air  gap  is  ~  X  .3132  = 
.0052.  Thus  the  armature  reluctance  is  so  small 
that  it  may  be  neglected.  The  permeance  of  the 
two  air  gaps  is  given  by  005a1xg  =  100  (about).  The 

coefficient  of  leakage  =  100+)250  =  3.5. 

As  the  coefficient  of  leakage  is  the  factor  used  in 
these  calculations,  the  permeance  of  the  leakage  paths 
is  the  desired  factor  for  its  determination.  In  the 
case  of  cylindrical  magnet  cores  parallel  to  each 
other,  they  are  obtained  from  Table  XIII.  given  in 
its  place  later.  It  is  thus  calculated  and  used. 
The  least  distance  separating  the  cores  (b)  is  di- 
vided by  the  circumference  of  a  core  (p)  giving  the 
ratio  (~)  of  least  distance  apart  to  perimeter  of  a 
core.  The  number  corresponding  in  columns  3  or  5 
is  multiplied  by  the  length  of  a  core.  The  product 
is  the  permeance.  Columns  2  and  4  give  the  re- 
luctance. To  reduce  to  average  difference  of  mag' 
netic  potential  divide  by  2. 


ELECTRO-MAGNETS,  DYNAMOS  AND  MOTORS.   91 

EXAMPLE. 

Calculate  the  permeance  between  the  legs  of  a 
magnet,  3  inches  in  diameter  and  12  inches  high 
and  5  inches  apart. 

Solution:  The  perimeter  =  3  X  3.14  =  9.42.  ^  = 

1  or  .5    nearly.     From  the  table    of  permeability 
we  find  6.278.     Multiplying  this  by   12  we  have 
6.278  X  12  =  75.336,  the  permeance.     Dividing  by 

2  we  have  ^f^  =  37.668,  the  permeance  for  use  in 
the  calculation  of  leakage  coefficient. 

It  will  be  observed  that  this  calculation  is  based 
entirely  on  the  ratio  stated,  and  that  absolute  di- 
mensions have  no  effect  on  it. 

For  flat  surfaces,  parallel  and  facing  each  other, 
the  following  method  precisely  comparable  to  the 
rule  for  specific  resistance  is  used: 

Rule  70.  The  permeance  of  the  air  space  between  flat 
parallel  surfaces  is  equal  to  their  average  area  multi- 
plied by  3.193  and  divided  by  their  distance  apart,  all 
in  inch  measurements. 

EXAMPLE. 

Determine  the  permeance  between  the  two  facing 
sides  of  a  square  cored  magnet  15  inches  long,  3 
inches  wide  and  8  inches  apart. 

Solution:  3  X  15  =  45  (the  average  area);  45  X 
3.193  -s-  8  =  17.96.  For  use  in  calculations  it 
should  be  divided  by  2  giving  8.98.  This  division 
by  2  is  to  reduce  it  to  the  average  difference  of  mag- 
netic potential  between  the  two  magnet  legs. 


92  ARITHMETIC  OF  ELECTRICITY. 

CALCULATIONS  FOR  MAGNETIC  CIRCUITS. 

A  magnetic  circuit  is  treated  like  an  electric  one. 
The  permeance  (analogue  of  conductance)  or  reluc- 
tance (analogue  of  resistance)  is  calculated  for  its 
four  parts,  magnet  core,  two  air  gaps,  and  armature 
core.  The  leakage  coefficient  is  determined  and  ap- 
plied. The  requisite  magnetizing  force  is  calcu- 
lated in  the  form  of  ampere  turns  (the  analogue  of 
volts  of  E.  M.  F.).  The  preceding  leakage  rules 
cover  the  case  of  parallel  leg  magnets.  For  others 
a  slight  change  is  requisite  in  the  leakage  calcula- 
tions, but  in  practice  an  average  can  generally  be 
estimated. 

EXAMPLE. 

Assume  the  magnet  and  armature  of  a  dynamo. 
The  magnet  is  of  cast  iron,  each  leg  is  cylindrical  in 
shape,  4  inches  in  diameter  and  20  inches  high. 
From  center  to  center  of  leg  the  distance  is  9  inches. 
The  armature  core  of  soft  wrought  iron  is  4  inches 
in  diameter  and  8  inches  long,  the  pole  pieces  curv- 
ing around  it  are  4  inches,  measured  on  the  curve 
inside,  by  8  inches  long.  The  air  gap  is  i  inch 
thick.  Calculate  the  reluctance  of  the  circuit  and 
the  ampere  turns  for  500,000  lines  of  force. 

Solution:  The  pole  pieces  approach  within  2$ 
inches  of  each  other.  This  leaves  l£  inches  of  the 
diameter  of  the  armature  core  embedded  or  included 
within  or  embraced  by  them.  One-half  of  this 


-ELECTRO-MAGNETS,  DYNAMOS  AND  MOTOES.   92 

amount  may  be  taken  and  added  to  2i  giving  3J  as 
the  average  depth  of  core  for  an  area  4  X  8  =  32 
square  inches.  The  lines  per  square  inch  of  arma- 
ture core  are  ^^  =  15,625  lines  per  square  inch. 
By  the  table  of  permeability,  4650  is  given  for  per- 
meability for  30,000  lines  in  soft  iron.  For  15,625 
lines  per  square  inch  9,000  can  safely  be  taken  for 
permeability.  Its  relative  reluctance  is  therefore 
32x^000  =  .000011  relative  armature  core  reluctance. 

(i) 

The  relative  reluctance  of  one  air  gap  (permeabil- 
ity =  1)  is  i  -*-  32  =  .0078  and  .0078  X  2  =  .0156  =s 
air  gaps  reluctance  (2). 

The  ratio  -  of  the  table  for  determining  the  leak- 
age between  cylindrical  magnet  legs  is  4  X53  ^  =  .4. 
5  is  the  distance  between  the  legs.  Permeance  cor- 
responding thereto  is  6.897,  which  multiplied  by  20, 
the  length  of  the  legs,  and  divided  by  2  for  average 
magnetic  potential  difference  gives  68.97  for  rela- 
tive effective  permeance  (3). 

The  relative  reluctance  of  the  air  gaps  and  arma- 
ture core  is  .015611;  the  reciprocal  or  permeance  is 
64.06  (4). 

For  coefficient  of  leakage  we  have  (64.06  -f-  68. 97) 
s-  64.06  =  2.08  (5). 

To  find  the  relative  reluctance  of  the  magnet  core 
whose  yoke  may  be  taken  as  of  mean  length  9  inches 


94  ARITHMETIC  OF  ELECTRICITY. 

and  of  area  equal  to  that  of  the  core  (3.14  X  &  — 
12.56)  we  have  to  first  determine  the  permeability. 
12'56  •  =  40,000  lines  per  square  inch,  corresponding 
to  a  permeability  of  258.  For  the  effective  reluc- 
tance of  the  magnet  core  introducing  the  factor  of 
leakage  (2.08)  we  have  the  expression  (20+122°5+X9)^82-08 
=  .0314. 

To  get  ampere  turns,  we  add  the  reluctances  of 
circuit,  multiply  by  .3132  and  by  the  required  lines, 
(.000011  +  .0156  +  .0314)  X  .3132  X  500,000  = 
7362  ampere  turns  required. 

In  the  above  calculations,  the  multiplication  by 
.3132  was  omitted  to  save  trouble,  relative  reluc- 
tances only  being  calculated,  until  the  end  when  one 
multiplication  by  .3132  brought  out  the  ampere 
turns.  The  leakage  appears  excessive  partly  be- 
cause of  the  high  reluctance  of  the  two  air  gaps. 
These  should  be  increased  in  area  and  reduced  in 
depth  if  possible.  The  leakage  is  also  high  on 
account  of  the  legs  of  the  magnet  being  close  to* 
gether.  Were  these  separated,  a  larger  armature 
core  might  be  used,  justifying  a  lower  speed  or  rota- 
tion of  armature,  reducing  reluctance  of  air  gaps  by 
increasing  their  area,  and  reducing  leakage  between 
magnet  legs  by  increasing  their  distance.  The 
magnet  legs  might  also  be  made  shorter,  thus  reduc- 
ing leakage. 


ELECTRO-MAGNETS,  DYNAMOS  AND  MOTORS.   95 

Thus  assume  the  magnet  core  of  the  same  cross- 
sectional  area,  but  only  10  inches  long  and  with  a 
distance  apart  of  legs  of  7  inches,  giving  a  7  X  10 
inch  armature  core  and  pole  piece  areas  (air  gap 
areas)  of  7  X  10  =  70  sq.  inches. 

For  leakage  ratio  we  have  (£)  =  ^  =  .56  giv- 
ing from  the  proper  table  6.000  (about),  *****  10 
=  30  relative  permeance  of  air  space  between  legs. 

For  air  gaps  reluctance  £  -5-  70  =  .00357  which 
for  the  two  gaps  gives  .00714  relative  reluctance. 

Treating  the  armature  core  as  a  prism  7  X  10  =  70 
sq.  inches  area  and  5  inches  altitude,  we  have  for 
lines  per  sq.  inch  500,000  *  70  =  7000  giving  it 
about  9000  and  reluctance  as  5  •*-  (70  X  9,000)  = 
.000008  reluctance. 

Air  gaps  and  armature  core  reluctance  »  .007143 

and  permeance  =  -^^  —  139. 


Coefficient  of  leakage  =  1-^±^)  =  1.21. 

If  the  depth  of  the  air  gaps  was  reduced  to  $•  inch 
the  coefficient  of  leakage  would  then  be  about  1.11. 

Every  surface  in  a  magnet  leaks  to  other  surfaces 
and  the  leakage  from  leg  to  leg  is  sometimes  but  one 
third  of  the  total  leakage.  In  practice  the  total 
leakage  often  runs  as  high  as  50#,  giving  a  coefficient 
of  2.00  and  in  other  cases  as  low  as  25#,  giving  a  co- 
efficient of  1.33. 


96  ARITHMETIC  OF  ELECTRICITY. 

DYNAMO  AKMATUKES. 

An  armature  of  a  dynamo  generally  comprises  two 
parts — the  core  and  the  winding.  The  core  is  of 
soft  iron.  Its  object  is  to  direct  and  concentrate 
the  lines  of  force,  so  that  as  many  as  possible  of 
them  shall  be  cut  by  the  revolving  turns  or  convolu- 
tions of  wire.  The  winding  is  usually  of  wire.  It 
is  sometimes,  however,  made  of  ribbon  or  bars  of 
copper.  Iron  winding  has  also  been  tried,  but  has 
never  obtained  in  practice.  The  object  of  the  wind- 
ing is  to  cut  the  lines  of  force,  thereby  generating 
electro  motive  force.  The  number  of  the  lines  of 
force  thus  cut  in  each  revolution  of  the  armature 
is  determined  from  the  intensity  of  the  field  per 
unit  area,  and  from  the  position,  area  and  shape  of 
the  armature,  coils  and  pole  pieces.  The  number 
thus  determined,  multiplied  by  the  number  of  times 
a  wire  cuts  them  in  a  second,  and  by  the  effective 
number  of  such  wires,  gives  the  basis  for  determin- 
ing the  voltage  of  the  armature. 

Rule  71.    One  volt  E.  OT.  F.  Is  generated  by  the  cutting 
of  108  (100,000,000)  lines  of  force  in  one  second. 

EXAMPLES. 

A  single  convolution  of  wire  is  bent  into  the  form 
of  a  rectangle  7  X  14  inches.  It  revolves  25  times  a 
second  in  a  field  of  20,000  lines  per  square  inch. 
What  E.  M.  F.  will  it  develop  at  its  terminals? 

Solution:  The  area  of  the  rectangle  is  7  X  14  =  98 


ELECTRO-MAGNETS,  DYNAMOS  AND  MOTORS.   97 

square  inches.  Multiplying  this  by  the  lines  of  force 
in  a  square  inch,  we  have  98  X  20,000  =  1,960,000. 
Each  side  of  the  rectangle  cuts  these  lines  twice  in 
a  revolution,  and  makes  25  revolutions  in  a  second. 
This  gives  25  X  2  X  1,960,000  =  98,000,000  lines 
cut  per  second,  corresponding  to  98  X  10*  X  10"*  = 
98  x  10~2  =  t&  volts  E.  M.  F.  generated,  or  4Afttfftft 
-  •£&  volts. 

The  field  of  the  earth  in  the  line  of  the  magnetic 
dip  —  .5  line  per  square  centimeter.  Calculate  a 
size,  number  of  layers,  and  speed  of  rotation  for  a 
one  volt  earth  coil. 

Solution  :  We  deduce  from  the  rule  the  following : 
Area  of  coil  X  revolutions  per  sec.  X  convolutions 
of  wire  X  .5  X  10~8  =  .5.  We  may  start  with  revolu- 
tions per  second,  taking  them  at  20.  Next  we  may 
take  50,000  convolutions.  20  x  50,000  X  .5  = 
500,000.  This  must  be  multiplied  by  200  to  give 
108;  in  other  words,  the  average  area  within  the 
wire  coils  must  be  100  square  centimeters,  or  10  X 
10  centimeters.  2  x  100  X  2000  X  500  X  .5  =  10", 
and  10"  X  10'8  =  1  volt. 

Rule  72.  The  capacity  «f  an  armature  for  current  is 
determined  by  tli  e  cross-section  of  its  conductors.  This 
should  be  such  as  to  allow  52O  square  mils  per  ampere 
=  1923  amperes  per  square  inch  area. 

EXAMPLE. 

A  drum  armature  coil  is  of  4  inches  diameter, 
and  is  wound  with  wire  -$fa  of  the  periphery  of  the 


98  ARITHMETIC  OF  ELECTRICITY. 

drum  in  diameter;  the  wire  is  100  feet  long.  Its 
E.  M.  F.  is  90  volts.  What  is  the  lowest  admissible 
external  resistance? 

Solution:  The  circumference  of  the  drum  is  3.14 
X  4  =  12.56  inches.  The  diameter  of  the  wire  is 
^  =  .0418  in.  or  42  mils.  The  area  of  the  wire  is 
2 12  X  3.14  =  1387  square  mils.  By  the  rule  the 
allowable  current  in  amperes  for  a  single  lead  of 
such  wire  is  *ffi  =  2.66  amperes.  But  on  a  drum 
armature  the  wire  lies  with  two  leads  in  parallel. 
Hence  it  has  double  the  above  capacity  or  2.66  X  2 
=  5.32  amperes.  The  resistance  of  such  wire  may 
be  taken  at  .137  ohm.  By  Ohm's  law  the  total  re- 
sistance for  the  current  named  must  be  ^  or  17 
ohms.  The  external  resistance  is  given  by  17  ^- 
.137  =  16.863  ohms. 

These  two  rules  enable  us  to  calculate  the  ca- 
pacity of  any  given  armature.  Certain  constants 
depending  on  the  type  of  armature  have  to  be  intro- 
duced in  many  cases. 

DRUM  TYPE  CLOSED  CIRCUIT  ARMATURES. 

For  these  armatures  the  following  rules  of  varia- 
tion hold,  when  they  do  not  differ  too  much  in  size, 
and  are  of  identical  proportions. 

Rule  73.  a.  The  E.  M.  F.  varies  directly  with  the 
•qu  arc  of  the  size  of  core  and  with  the  number  of  turns 
of  wire. 


ELECTRO-MAGNETS,  DYNAMOS  AND  MOTORS.   99 

b.  The  current  capacity  varies  with     the    sixth    root 
of  the  size  of  core  for  identical  !•:.  M.  F. 

c.  The  resistance  varies  directly  with  the  cube  of  the 
number  of  turns  and  inversely  with  the  size  of  core. 

d.  The  amperage  on  short  circuit  varies  directly  with 
the  cube  of  the  size  and  inversely  with  the  square  ot 
the  number  of  turns. 

In  these  rules  the  proportions  of  the  drum  are 
supposed  to  remain  unchanged.  Size  may  be  re- 
ferred to  any  fixed  factor  such  as  diameter,  as  lineal 
size  is  referred  to. 

These  rules  enable  us  to  calculate  an  armature 
for  any  capacity  and  voltage.  As  a  starting  point  a 
given  intensity  of  field,  speed  of  rotation,  and  num- 
ber of  turns  of  wire  and  size  of  wire  has  to  be  taken. 
The  wire  is  selected  to  completely  fill  the  periphery 
of  the  drum.  Then  a  trial  armature  is  calcu- 
lated of  the  required  voltage  and  its  amperage  is 
calculated.  With  this  as  a  basis,  by  applying  Eule 
73,  sections  a  and  #,  the  size  of  an  armature  for  the 
desired  current  capacity  is  calculated,  the  E.  M.  F. 
being  kept  identical.  As  a  standard  for  medium 
sized  machines  20£  of  the  turns  of  wire  may  be  con- 
sidered inactive. 

EXAMPLE. 

Calculate  a  100  volt,  20  ampere  armature,  whose 
length  shall  be  twice  its  diameter,  to  work  at  a 
speed  of  15  revolutions  per  second. 

Solution:  Take  as  intensity* of  field  20,000  lines 
per  square  inch.  Allow  80£  of  active  turns  of 


100  ARITHMETIC  OF  ELECTRICITY. 

wire.  Start  with  a  core  8  X  16  =  128  square  inches, 
including  128  X  20,000  =  256  X  10*  lines  of  force. 
The  given  speed  is  15  rotations  per  second.  For 
the  number  of  active  turns  of  wire  per  volt  we 
have  to  divide  108  or  100,000,000  by  one  half  the 
lines  of  force  cut  by  one  wire  per  second.  This 
number  is  256  X  104  X  15,  or  38,400,000;  and 

^JoSw  =  2'6  turns-  For  10°  volts>  therefore,  260 
active  turns  are  needed.  If  one  half  the  lines  were 
not  taken  the  result  would  be  one  half  as  great,  be- 
cause each  line  cuts  each  line  of  force  twice  in  a 
revolution,  and  in  the  computation  a  single  cutting 
per  revolution  only  is  allowed  for. 

The  reason  for  thus  taking  one  half  the  lines  cut 
by  a  single  wire  as  a  base  is  because  in  the  drum  arma- 
ture the  wires  work  in  two  parallel  series,  giving  one 
half  the  possible  voltage.  The  actual  turns  are 
260  •*-  .80  =  325,  say  324  turns.  Assume  it  to  be 
laid  in  two  layers  giving  162  turns  to  the  layer. 
The  space  occupied  by  a  wire  is  equal  to  the  peri- 
meter divided  by  the  number  of  wires  or  T¥?  =  .154 
in.  Allowing  25#  for  thickness  of  insulation,  lost 
space,  etc.,  we  have  .115  in.  or  115  mils  as  the 
diameter  of  the  wire.  In  the  drum  armature  as  just 
stated  the  wire  is  parallel,  so  that  the  area  of  one 
lead  of  wire  has  to  be  doubled,  giving  10,573  X 
2  square  mils  as  the  area  of  the  two  parallel  leads. 
This  is  enough  for  40  amperes  or  double  the  amper- 
age required.  This  capacity  is  reached  by  taking 


ELECTRO-MAGNETS,  D?-NA#f)A.A&J)WOrOSc,.  101 

520  square  mils  per  ampere  as  the  proper  cross-sec- 
tional area  of  the  wire.  (Rule  72.) 

We  must  therefore  reduce  the  size  to  give  ^  the 
ampere  capacity;  this  reduction  (by  Rule  73  Z>)  is  in 
the  ratio  1  :  j£*  ~  1  :  .89018  ;  the  size  therefore  is 
8  X  .89018  diameter  by  16  X  .89018  length  =  7.12 
X  14. 24  inches. 

Applying  a  for  voltage  we  have  for  the  same  num- 
ber of  turns  on  the  new  armature  a  voltage  in  the 
ratio  of  1  :  .890182  or  about  &  of  that  required.  We 
must  therefore  divide  the  number  of  turns  in  the 
trial  armature  by  .890182,  giving  for  the  number  of 
turns  H^  =  409,  say  410  turns. 

To  prove  the  operation  we  first  determine  the 
voltage  of  the  new  armature.  Its  area  is  7.12  X 
14.24  =  101.4  square  inches  including  2,028,000  lines 
of  force.  The  active  wires  are  410  X  .8  =  328. 
We  have  for  the  voltage  =  .^,000^1^x15  =  99>78 

volts. 

The  relative  capacity  of  the  wire  is  deduced  from 
the  square  of  its  diameter.  The  circumference  of 
the  new  armature  is  7.12  X  3.14  =  22.3568.  There 
are  205  turns  in  a  layer  giving  as  diameter  of  wire 
^^  =  .1091  mils.  This  must  be  squared,  giving 
.0119,  and  compared  with  the  square  of  the  corre- 
sponding number  for  the  original  armature.  This 
number  was  25  -*•  .162  =  .154  inch.  .1542  =  .02371 


102  A?1T^'IFT-:C  OF  ELECTRICITY. 

and  .01190  •*-  .02371  =  i  (nearly),  showing  that 
the  new  armature  has  one  half  the  ampere  capac- 
ity of  the  old,  or  40  X  |  =  20  amperes  as  re- 
quired. 

The  gauge  of  the  wire  is  reached  by  making  the 
same  allowance  for  insulation  and  lost  space,  viz., 
25#.  .1091  X  .75  =  .0818  in.  or  81.8  mils  diameter, 
for  size  of  wire.  Of  course  there  is  nothing  absolute 
about  25#  as  a  loss  coefficient;  it  will  vary  with 
style  of  insulation  and  even  to  some  extent  with  the 
gauge  of  wire.  But  as  Eule  73  is  based  upon  the 
assumption  that  this  loss  is  a  constant  proportion  of 
the  diameter  of  the  wire,  too  great  a  variation  of 
sizes  should  not  be  allowed  in  its  application.  In 
other  words  the  trial  armature  should  be  as  near  as 
possible  in  size  to  the  final  one. 

Suppose  on  the  other  hand  that  an  armature  for 
100  amperes  was  required.  This  is  for  2^  times 
40  amperes  (the  capacity  of  the  first  calculated  or 
trial  armature). 

Applying  b  we  extract  the  6th  root  of  2^. 
=  1.1653  (by  logarithms  or  by  a  table  of  6th 
roots).  The  size  of  the  new  armature  is  therefore 
8  X  1.1653  by  16  X  1.1653  or  9.3224  X  18.6448 
inches. 

Applying  a  for  voltage  we  have  for  the  same  num- 
ber of  turns  of  the  new  armature  a  voltage  in  the 
ratio  of  1.16532  :  1  or  1.358  times  too  great.  We 


ELECTRO-MAGNETS,  DYNAMOS  AND  MOTORS.  103 

must  therefore  multiply  the  original  turns  by  the 
reciprocal  of  1.358,  giving  ^  =  239  turns. 

To  prove  the  voltage,  we  multiply  239  by  .8  for 
the  active  turns  of  wire,  giving  191.2  turns.  The 
area  of  the  armature  is  9.32  X  18.64  =  172.7  square 
inches.  For  voltage  this  gives  172.7  X  191.2  X  20,- 
000  X  15  X  10-8  =  98.6  volts  (about). 

To  prove  the  capacity  we  must  divide  the  circum- 
ference of  the  new  armature,  9.32  X  3.14  =  29.26 
inches,  by  the  turns  of  wire  in  one  layer,  4*  =  120 
turns  (about).  This  gives  a  diameter  of  244  mils 
(nearly).  The  ratio  of  capacities  of  the  original  and 
this  wire  is  .2442  -5-  .1542  inches  =  .059536  -*-  .02371 
=2.51  corresponding  to  40  X  2.51  =  100  amperes. 

These  results,  owing  to  omissions  of  decimals,  do 
not  come  out  exactly  right  and  it  is  quite  unneces- 
sary that  they  should.  The  accuracy  is  ample  for 
all  practical  purposes.  For  armature  dimensions  it 
would  be  quite  unnecessary  to  work  out  to  the  sec- 
ond decimal  place.  It  would  answer  to  take  as  ar- 
mature sizes  in  the  two  cases  given  7  X  14^  inches 
and  9^  X  18#  inches. 

It  is  also  to  be  noted  that  a  very  low  rate  of  rota- 
tion was  taken.  25  to  30  turns  per  second  would 
not  have  been  too  much.  The  latter  would  give 
double  -the  voltage  and  the  same  amperage. 

FIELD  MAGNETS  OF  DYNAMOS. 
The  calculation  for  a  magnetic  circuit  given  on 


104  ARITHMETIC  OF  ELECTRICITY. 

pages  92  et  seq.,  is  intended  to  supply  an  example  of 
the  calculation  of  the  circuit  formed  by  a  field  mag- 
net and  its  armature,  such  as  required  for  dynamos. 
The  leakage  of  lines  of  force  is  and  can  only  be  so 
incompletely  calculated  that  it  is  probably  the  best 
and  most  practical  plan  to  assume  a  fair  leakage 
ratio  and  to  make  the  magnet  cores  larger  than  re- 
quired by  the  lines  of  force  of  the  armature  in  this 
ratio.  (  A  low  multiplier  to  adopt  is  1.25,  which  is 
lower  than  obtains  in  most  cases;  1.50  is  probably  a 
good  average. 

Rule  74.  The  cross-sectional  area  of  the  field-magnet 
cores  is  equal  to  the  lines  of  force  in  the  field  divided  by 
the  magnetic  flux  (column  B)  for  the  material  selected 
and  corresponding  to  the  chosen  permeability  GU.),  mul- 
tiplied by  the  leakage  coefficient. 

A  good  range  for  permeability  is  from  200  to  400 
giving  for  wrought  iron  from  100,000  to  110,000 
lines  of  force  per  square  inch  and  for  cast  iron  from 
35,000  to  45,000  lines  per  square  inch;  for  the  field 
from  15,000  to  20,000  lines  per  square  inch  may  be 
taken. 

The  permeability  table  gives  data  for  different 
qualities  of  iron. 

EXAMPLE. 

Taking  the  100  volt  100  ampere  armature  last  cal- 
culated, determine  the  size  of  field-magnet  cores  to 
go  with  it,  and  the  ampere  turns  and  other  data. 

Solution:  Assume  20, 000  lines  of  force  per  square 


ELECTRO-MAGNETS,  DYNAMOS  AND  MOTORS.  105 

inch  in  the  field,  45,000  in  cast  iron  and  110,000  in 
wrought  iron  core  and  a  leakage  coefficient  of  1.25. 
We  have  for  total  lines  of  force  passing  through 
armature  172.7  X  20,000  =  3,454,000;  cross-sectional 
area  for  cast  iron  core  ^^°  X  1.25  =  96  square 
inches;  cross-sectional  area  for  wrought  iron  core 
X  1.25  =  39  square  inches. 


As  length  of  cores  we  may  take  20  inches  with 
a  distance  between  them  of  10  inches.  Assume 
wrought  iron  to  be  selected.  If  cylindrical  they 
would  be  7  inches  in  diameter  to  give  the  required 
cross-sectional  area.  The  yoke  connecting  them 
would  average  in  length  10  +  7  =  17  inches,  giving 
for  magnet  cores  and  yoke  a  length  of  17  +  20  +  20 
=  57  inches.  The  reluctance  of  cores  and  yoke 
(Rule  68)  -  Vx'aro  (taking  A«  =  200)  which  reduces 
to  .00132  (1). 

The  armature  area  is  172.7  inches.  As  average 
length  of  the  path  of  lines  of  force  through  it  5 
inches  may  be  taken.  As  it  passes  only  20,000  lines 
of  force  per  square  inch  of  field  its  permeability  is 
high,  say  9000.  Its  reluctance  is  given  by  ^^'Soo* 

This  is  so  low  that  it  may  be  neglected. 

The  area  of  each  air-gap  may  be  taken  as  173 
square  inches,  and  of  depth  of  two  windings  plus 
about  -rV  inch  for  clearance  or  windage  giving 
(.224  X  2)  +  .1  =  about  .6  inch  for  its  depth.  Its 


106  ARITHMETIC  OF  ELECTRICITY. 

reluctance  is  ^-~^=*  .00108.     As  there  are  two  air 

17o 

gaps  we  may  at  once  add  their  reluctances  giving 
.00216(2).  " 

By  Rule  67  the  ampere  turns  are  equal  to  the 
product  of  the  reluctances  (1)  and  (2),  by  the  lines 
of  force  giving  (.00121  +  .00216)  X  (172.7  +  20,000) 
=  11640  ampere  turns. 

The  proper  size  of  wire  for  series  winding  may  be 
determined  by  Sir  William  Thompson's  rule  that  in 
series  wound  dynamos  the  resistance  of  the  field  mag- 
net windings  should  be  &  that  of  the  armature.  The 
length  of  the  wire  in  the  armature  is  equal  approxi- 
mately, to  the  circumference  9.32  X  3.14  =  29.26 
multiplied  by  the  number  of  turns  (240)  giving 
29.26  X  240  =  7022  inches. 

The  wire  turns  on  the  field  magnets  are  found  by 
dividing  the  ampere  turns  by  the  amperes  giving 
^W4  =  232  turns.  The  circumference  of  the  mag- 
net leg  is  7.0  X  3.14  =  22  inches.  The  total  length 
of  wire  is  therefore,  approximately,  232  X  22  =  5104 
inches. 

To  compare  the  resistances  we  must  use  *^P  for 
the  length  of  the  armature  wire,  because  it  is  in 
parallel,  and  therefore  is  \  the  length  and  j  the  re- 
sistance of  the  full  wire  in  one  length.  Dividing  by 
4  introduces  this  factor. 

As  the  resistances  of  the  wires  are  to  be  in  the  ratio 
of  2  :  3,  we  have  by  Rule  13  (calling  the  thickness 
of  armature  wire  244  X  .75  =  183  mils  to  allow  for 


ELECTRO-MAGNETS,  DYNAMOS  AND  MOTORS.  107 

insulation,  etc.),  2:3  ::  1832  X  5104  :  z2  X  **£*-,  and 
solving  we  find  x>  =  73026  .  •.  x  =  270  mils. 

For  shunt  winding  Sir  William  Thompson's  rule 
is  that  the  product  of  armature  and  field  resistance 
should  equal  the  square  of  the  external  resistance. 
The  latter  may  be  taken  (Ohm's  law)  as  equal  to 
=  l  ohm-  Fr°Perly  the  armature  resist- 


ance should  be  allowed  for,  but  it  is  so  small  that  it 
need  not  be  included.  We  have  therefore,  arma- 
ture resistance  X  field  resistance  =  I2  X  1. 

The  armature  resistance  is  .0419  ohms.  There- 
fore the  field  resistance  is  -^gj  =  24  ohms.  The  cur- 
rent through  this  is  equal  to  *$£  =  4  amperes 
(nearly).  Therefore  n|39  =  2910  turns  of  wire  are 
needed.  The  length  of  such  wire  will  be  ^  ^^ 
=  5335  feet.  The  resistance  is  about  4.4  ohms  per 
1000  feet  corresponding  to  about  .48  mils  diameter. 

THE  KAPP  LINE. 

Mr.  Gisbert  Kapp,  C.  E.  who  has  given  much  in- 
vestigation to  the  problems  of  the  magnetic  circuit 
and  especially  to  dynamo  construction,  is  the  orig- 
inator of  this  unit.  He  considered  the  regular  C. 
G.  S.  line  of  force  to  be  inconveniently  small.  He 
adopted  as  a  line  of  force  the  equivalent  of  6000  C. 
G.  S.  lines  and  as  the  unit  of  area  one  square  inch. 
Therefore  to  reduce  Kapp  lines  to  regular  lines  of 
force  they  must  be  multiplied  by  6000,  and  ordi- 


108  ARITHMETIC  OF  ELECTRICITY. 

nary  lines  of  force  must  be  divided  by  6000  to  obtain 
Kapp  lines.  These  lines  are  often  used  by  English 
engineers.  The  regular  system  is  preferable  and  by 
notation  by  powers  of  ten  can  be  easily  used  in  all 
cases. 


CHAPTER  X. 

ELECTRIC  RAILWAYS. 

SIZES  OF  FEEDERS. 

To  calculate  the  sizes  of  feeders  for  a  trolley  line 
Eules  23,  24,  and  25  in  Chapter  V.  will  be  found  use- 
ful in  conjunction  with  the  following  ones: 

A.  For  load  at  end  of  feeders: 

Rule  7o.  The  cross-section  of  the  feeder  in  cir- 
cular mils  is  equal  to  the  product  of  1O.79  times  the 
current  in  amperes  times  the  length  of  the  con- 
dnctor  in  feet,  divided  by  the  allowable  drop  in 
volts.  . 

EXAMPLE. 

What  should  be  the  cross  section  of  a  feeder  3,000 
feet  long  carrying  90  amperes  with  35  volts  drop? 

Solution:  3,000  X  90  X  10.79  =  2,913,300.  Di- 
viding this  by  35  gives  83,237  circular  mils.  This 
would  correspond  to  a  No.  1  wire,  which  has  83,694 
cir.  mils  area. 

All  computations  of  this  kind  should  be  checked 
by  table  XVI.  of  current  capacity  on  page  153  in 
order  to  be  sure  that  the  wire  will  not  become  heated 
above  the  allowable  limit. 


110  ARITHMETIC  OF  ELECTRICITY. 

Beferring  the  above  example  to  this  table,  it  is 
seen  that  a  No.  1  wire  will  carry  80  amperes  with  a 
rise  in  temperature  of  18°  F.,  and  110  amperes  with 
a  rise  of  36°  F.  Hence  the  current  of  90  amperes 
will  cause  a  rise  of  24°  F.  This  is  calculated  by 
simple  proportion;  subtracting  80  from  90  and  from 
110  gives  10  and  30  as  the  respective  differences  and 
shows  90  to  lie  at  just  one-third  the  distance  from 
80  to  110.  Hence  the  resulting  temperature  rise 
will  be  at  one-third  the  distance  between  18°  and 
36°.  The  difference  between  these  last  two  figures 
is  18°,  one-third  of  which  is  6°.  Add  this  6°  to 
18°  gives  us  24°  F.  as  the  answer. 

It  is  often  desirable  to  compute  the  drop  on  a 
feeder  carrying  a  given  current;  this  is  done  by  the 
following : 

Rale  76.  The  drop  in  volts  on.  any  conductor  is 
found  by  multiplying  together  1O.79,  the  current  in 
amperes  and  its  length  in  feet,  then  divide  this 
product  by  its  area  in  circular  mils. 

EXAMPLE. 

What  is  the  drop  on  a  feeder  2,800  feet  long,  of 
105,592  cir.  mils  area  and  carrying  a  current  of  125 
amperes  ? 

Solution:  10.79  X  125  X  2,800  =  3,776,500. 

Dividing  this  product  by  105,592  gives  35.76 
volts  drop. 

B.  For  a  uniformly  distributed  load: 


ELECTRIC  RAILWAYS.  Ill 

The  effect  of  a  uniform  distribution  of  load  along 
a  main  or  trolley  wire  is  the  same  as  that  of  half 
the  total  current  passing  the  full  length  of  the  wire; 
hence  we  require  but  half  the  cross-section  needed  to 
deliver  the  current  at  the  extremity  of  the  wire. 

This  is  readily  done  by  substituting  the  constant 
5.4  in  place  of  10.79  in  the  foregoing  rules. 

In  designing  electric  railway  circuits  where  the 
track  forms  the  path  for  the  return  current  the  rails 
should  be  of  ample  area  and  well  bonded,  with  an 
extra  bare  wire  connected  to  the  bonds  and  materially 
reducing  the  drop  in  the  track  circuit 

POWER  TO  MOVE  CABS. 

At  ordinary  speeds  on  a  level  track  in  average  con- 
dition it  is  safe  to  assume  that  the  force  necessary 
to  move  a  car  is  30  pounds  per  ton  of  weight  of  car. 

Rale  77.  To  find  the  force  required  to  pull  or 
push  a  car  on  a  level  track  in  average  condition, 
multiply  the  weight  of  the  car  in  tons  by  3O. 

EXAMPLE. 

Find  the  force  required  to  drag  a  car  weighing  7 
tons  on  a  level  track. 

Solution:  7  X  30  =  210  Ibs.     Ans. 

Should  it  be  required  to  find  the  force  needed  to 
start  a  car  on  a  level,  or  to  propel  it  when  round- 
ing a  curve,  substitute  the  constant  70  in  place  of 
30  in  the  foregoing  rule. 


112  ARITHMETIC  OF  ELECTRICITY. 

EXAMPLE. 

What  force  is  needed  to  start  an  8  ton  car  on  a 
level  track? 

Solution:  8  X  70  =  560  pounds.    Ans. 

As  the  above  does  not  take  into  account  the  speed 
of  the  car  we  shall  have  to  add  this  factor  in  order 
to  find  the  horse  power  needed  to  move  it;  we  will 
also  allow  for  the  efficiency  of  the  motors. 

Rule  78.  To  find  the  horse  power  required  to  move 
a.  car  along?  a  level  track  multiply  together  the  dis- 
tance in  feet  traveled  per  minute  and  the  force  in 
pounds  necessary  to  move  the  car  (as  found  by  Rule 
77),  and  divide  the  result  by  33,OOO  times  the  ef- 
ficiency of  the  motors* 

EXAMPLE. 

What  horse  power  is  needed  to  propel  a  loaded  car 
weighing  9  tons  along  a  level  track  at  the  rate  of 
800  feet  per  minute,  with  motors  of  70  per  cent, 
efficiency  ? 

Solution :  Force  to  move  car  is  9  X  30  =  270 
pounds.  The  product  of  800  X  270  =  216,000  foot 
pounds  per  minute.  Dividing  this  by  33,000  gives 
6.54  H.  P.  required  to  propel  the  car.  Dividing  by 
the  efficiency  .70  gives  9.34  H.  P.  to  be  delivered  to 
the  motors. 

It  will  be  noted  in  this  solution  that  the  quantity 
216,000  should,  according  to  the  rule,  have  been  di- 
vided by  the  product  of  33,000  times  .70;  it  was, 


ELECTRIC  RAILWAYS.  113 

however,  divided  by  these  two  factors  successively  in 
order  to  show  the  difference  between  the  power  actu- 
ally moving  the  car  and  that  supplied  to  the  motors. 

In  computing  the  power  taken  by  a  car  ascending 
a  grade  the  equivalent  perpendicular  rise  of  the  car 
together  with  its  weight  in  pounds  have  to  be  consid- 
ered in  addition  to  the  factors  involved  in  the  rule 
just  preceding. 

Rule  79.  To  find  the  horse  power  required  to  pro- 
pel a  car  up  a  grade,  take  the  product  of  the 
perpendicular  distance  in  feet  ascended  by  the  car 
in  one  minute  multiplied  by  its  freight  in  pounds; 
to  this  add  the  product  of  the  horizontal  distance 
in  feet  traveled  in  one  minute  multiplied  by  the 
force  in  pounds  required  to  propel  the  car;  divide 
this  sum  by  33,OOO  times  the  efficiency  of  the  mo- 
tors. 

Note. — The  grade  of  a  road  or  track  is  generally 
stated  as  being  so  many  per  cent.  This  means  that 
for  any  given  horizontal  travel  of  a  car  its  change  of 
altitude  when  referred  to  a  fixed  horizontal  plane  is 
expressed  as  a  certain  percentage  of  the  horizontal 
travel.  For  illustration;  if  a  car  while  traveling 
horizontally  100  feet  has  a  total  vertical  rise  (or 
fall)  of  7  feet,  the  incline  on  which  it  moves  is 
termed  a  7  per  cent  grade. 

EXAMPLE. 

Find  the  electrical  horse  power  taken  by  the  motors 
of  an  8  ton  car  to  propel  it  up  a  5  per  cent  grade  at 


114  ARITHMETIC  OF  ELECTRICITY. 

a  speed  of  1,000  feet  per  minute,  the  motors  having 
70  per  cent  efficiency. 

Solution:  Perpendicular  rise  of  car  is  1,000  feet 
X  -05  =  50  feet.  Weight  of  car  in  pounds  is  8  X 
2,000  =  16,000  pounds.  Product  of  lift  and  weight 
is  50  X  16,000  =  800,000  foot  pounds.  Force  re- 
quired to  propel  car  is  30  X  8  =  240  pounds.  Pro- 
duct of  force  and  distance  is  240  X  1,000  =240,000 
foot  pounds.  Sum  of  the  two  products  is  800,000  + 
240,000  =  1,040,000  total  foot  pounds. 

Product  of  33,000  by  efficiency  is  33,000  X  .70  = 
23,100.  Electrical  H.  P.  is  the  quotient  of  1,040,000 
-f-  23,100  =  45.021  H.  P.  Ans. 


CHAPTEE  XL 

ALTERNATING    CURRENTS. 

By  far  the  greater  part  of  calculations  in  the  do- 
main of  alternating  currents  lie  in  the  realm  of 
trigonometry  and  the  intricacies  of  the  calculus. 
On  this  account  it  is  hoped  that  the  following  pres- 
entation of  some  of  the  simpler  formulae  may  prove 
welcome  to  the  craft. 

A  current  flowing  alternately  in  opposite  directions 
may  be  considered  as  increasing  from  zero  to  a 
certain  amount  flowing  in,  say,  the  positive  direc- 
tion, then  diminishing  to  zero  and  increasing  to  an 
equal  amount  flowing  in  the  negative  direction  and 
again  decreasing  to  a  zero  value.  This  action  is 
repeated  indefinitely.  The  sequence  of  a  positive 
and  negative  current  as  just  described  is  called  a 
cycle. 

The  frequency  of  an  alternating  current  is  the 
number  of  cycles  passed  through  in  one  second.  An 
alternation  is  half  a  cycle.  That  is  to  say,  an  alter- 
nation may  be  taken  as  either  the  positive  or  the 
negative  wave  of  the  current. 


116  ARITHMETIC  OF  ELECTRICITY. 

The  frequency  may  be  expressed  not  only  in  cycles 
per  second  but  in  alternations  per  minute. 

Since  one  cycle  equals  two  alternations  we  can 
interchange  these  expressions  as  follows: 

Rule  8O.  A.  Having  given  the  cycles  per  second, 
to  find  the  alternations  per  minute  multiply  the 
cycles  per  second  by  12O.  B.  Having  given  the  al- 
ternations per  minute,  to  find  the  cycles  per  second 
divide  the  alternations  per  minute  by  12O. 

EXAMPLES. 

If  a  current  has  60  cycles  per  second,  how  many 
alternations  are  there  per  minute? 

Solution:  60X120  —  7,200  alternations. 

A  current  has  15,000  alternations  per  minute; 
how  many  cycles  per  second  are  there  ? 

Solution:  15,000-^120  =  125  cycles  per  second. 

A  bipolar  dynamo  having  an  armature  with  but 
a  single  coil  wound  upon  it  (like  an  ordinary  mag- 
neto generator)  gives  one  complete  cycle  of  current 
for  every  revolution  of  the  armature.  That  is  to 
say,  its  frequency  equals  the  number  of  revolutions 
per  second.  A  four-pole  generator  will  have  a  fre- 
quency equal  to  twice  the  revolutions  per  second,  etc. 

Rule  81.  To  find  the  frequency  of  any  alternator, 
divide  the  revolutions  per  minute  by  6O  and  multiply 
the  quotient  by  the  number  of  pairs  of  poles  in  the 
field. 

EXAMPLE. 

Find  the  frequency  of  a  16-pole  alternator  run- 
ning at  937.5  revolutions  per  minute. 


ALTERNATING  CURRENTS.  117 

Solution:  937.5  -r-  60  =  15.625  rev.  per  second. 
15.625  X  8  =  frequency  of  125  cycles  per  second. 

Electrical  measuring  instruments  used  on  alternat- 
ing currents  do  not  indicate  the  maximum  volts  or 
amperes  of  such  circuits,  but  the  effective  values  are 
what  they  show.  These  effective  values  are  the  same 
as  those  of  a  continuous  current  performing  the 
same  work. 

Rule  82.  The  maximum  volts  or  amperes  of  an 
alternating:  current  may  be  found  by  '  multiplying: 
the  average  volts  or  amperes  by  1.11.  Reciprocally, 
the  average  values  can  be  found  by  taking;  .707 
times  the  maximum  values. 

Xote  that  these  figrures  are  strictly  true  only  for 
an  exactly  sinusoidal  current. 

EXAMPLE. 

Find  the  maximum  pressure  of  an  alternating 
current  of  55  volts. 

Solution:  55  X  1.11  =  61.05  volts.    Ans. 

SELF-INDUCTION. 

In  an  alternating  current  circuit  the  flow  of  a 
current  under  a  given  voltage  is  determined  not 
only  by  the  resistance  of  the  conductor  in  ohms  but 
also  by  the  self-induction  of  the  circuit.  Suppose  a 
current  to  start  at  zero  and  increase  to  10  amperes  in 
a  coil  of  1,000  turns  of  wire.  This  magnetizing 
force,  growing  from  zero  to  10,000  ampere-turns, 
surrounds  the  coil  with  lines  of  force  whose  action 


118  ARITHMETIC  OF  ELECTRICITY. 

upon  the  current  in  the  coil  is  such  as  to  resist  its 
increase.  Conversely,  when  the  current  is  decreas- 
ing from  10  amperes  to  zero,  the  lines  of  force 
change  their  direction  and  tend  to  prolong  the  flow 
of  current.  This  opposing  effect  which  acts  on  a 
varying  or  an  alternating  current  is  caller  the  counter 
E.  M.  F.  or  E.  M.  F.  of  self-induction  and  is  meas- 
ured in  volts. 

Rule  83.  The  K.  M.  F.  of  self-induction  of  a  given 
coil  is  found  by  multiplying  tog-ether  12.5664,  the 
total  number  of  turns  in  the  coil,  the  number  of 
turns  per  centimeter  length  of  the  coil,  the  sec- 
tional area  of  the  core  of  the  coil  in  square  centi- 
meters, the  permeability  of  the  magnetic  circuit 
and  the  current;  divide  the  resulting  product  by 
1,OOO,OOO,OOO,  multiplied  by  the  time  taken  by  the 
current  to  reach  its  maximum  value. 

EXAMPLE. 

Find  the  volts  of  counter  E.  M.  F.  in  a  coil  of  300 
turns  wound  uniformly  on  a  ring  made  of  soft  iron 
wire,  the  ring  having  a  mean  circumference  of  60 
centimeters  and  an  effective  sectional  area  of  25 
square  centimeters;  its  permeability  to  be  taken  as 
200,  and  a  current  of  5  amperes  in  the  coil  requires 
.02  second  to  reach  its  maximum  value. 

Solution:  Product  of  12.5664  X  300  X  5  X  25 
X  200  X  5  is  471,240,000.  Product  of  10°  X  -02  is 
20,000,000.  Dividing  the  former  product  by  the 
latter  gives  23.562  volts,  Answer. 

The  coefficient  of  self-induction  or,  as  it  is  more 


ALTERNATING  CURRENTS.  119 

frequently  termed,  the  inductance  of  a  coil,  is  meas- 
ured by  the  number  of  volts  of  counter  E.  M.  F. 
when  the  current  changes  at  the  rate  of  one  ampere 
per  second.  (Atkinson.) 

The  unit  of  inductance  is  the  henry. 

Rule  84.  The  inductance  of  a  coil  is  found  by 
multiplying?  tog-ether  12.5664,  the  total  number  of 
turns  in  the  coil,  the  number  of  turns  per  centimeter 
lens th,  the  sectional  area  of  the  core  of  the  coil  in 
square  centimeters,  and  the  permeability  of  the 
magnetic  circuit;  divide  the  resulting;  product  by 
1,000,000,000. 

EXAMPLE. 

Find  the  inductance  of  the  coil  specified  in  the 
preceding  example. 

Solution:  The  factors  are  the  same  as  before, 
omitting  the  current  and  time.  Product  of  12.5664  X 
300  X  5  X  25  X  200  is  94,248,000.  Dividing  this 
by  1,000,000,000  gives  the  inductance  .094248 
henrys. 

The  E.  M.  F.  of  self-induction  may  be  computed 
when  the  inductance,  the  current  and  the  time  taken 
for  the  current  to  reach  its  maximum  are  known. 

Rule  85.  To  find  the  E.  M.  F.  of  self-induction 
divide  the  product  of  the  inductance  and  current  by 
the  time  of  current  rise. 

EXAMPLE. 

Using  again  the  data  of  the  foregoing  examples, 
find  the  counter  E.  M.  F.,  the  inductance  being 


120  ARITHMETIC  OF  ELECTRICITY. 

.094248  henrys,  the  current  5  amperes,  and  the  time 
.02  second. 

Solution:  5  X  .094248  =  .47124;  dividing  this 
by  .02  gives  23.562  volts,  as  before. 

Rule  SO.  The  resistance  due  to  self-induction 
equals  6.2S32  times  the  product  of  the  frequency 
and  the  inductance. 

EXAMPLE. 

Find  the  inductive  resistance  of  a  circuit  whose 
frequency  is  60  cycles  per  second  and  the  inductance 
is  .05  henry. 

Solution:  6.2832  X  60  X  -05  =  18.8496  ohms. 
Ans. 

The  time  constant  of  an  inductive  circuit  is  a 
measure  of  the  growth  or  increase  of  the  current. 
It  is  the  time  required  by  the  current  to  rise  from 
zero  to  its  average  value.  The  average  value  of  an 
alternating  current  is  .634  times  its  maximum  value. 
It  must  not  be  confused  with  its  effective  value, 
which  is  .707  times  the  maximum. 

The  average  value  may  be  obtained  by  multiplying 
the  effective  value,  as  shown  by  instruments,  by  .897. 

Rule  87.  To  find  the  time  constant  of  a  coil  or 
circuit,  divide  its  inductance  by  its  resistance. 

EXAMPLE. 

What  is  the  time  constant  of  a  coil  whose  induct 
ance  is  3.62  henrys  and  resistance  is  .20  ohms. 
Solution:  3.62  -~  20  =  .181  second.    Ans. 


CHAPTEE  XII. 

CONDENSERS. 

A  condenser,  though  it  will  allow  no  current  to 
pass  through  it,  yet  it  will  accumulate  or  store  up 
a  quantity  of  electricity  depending  on  various  factors 
which  the  following  rules  will  show: 

Rule  SS.  The  quantity  stored  equals  the  product 
of  the  K.  M.  F.  applied  and  the  capacity  of  the 
condenser. 


Rule   89.    The    capacity   of   a   condenser   equals    the 
quantity  stored  divided  by  the  applied  E.  M.  F. 


E 

Rule  9O.    The  E.  M.  F.  applied  to  a  condenser  equals 
the  quantity  stored  divided  by  its  capacity. 


The  quantity  stored  in  a  condenser  is  measured 
in  coulombs  (i.  e.,  ampere-seconds)  ;  the  E.  M.  F. 
in  volts,  and  the  capacity  in  farads.  Condensers  in 
practical  use  have,  however,  so  small  a  capacity  that 


122  ARITHMETIC  OF  ELECTRICITY. 

it  is  usually  stated  in  microfarads  and  the  quantity 
in  microcoulombs. 

EXAMPLES. 

A  battery  of  30  volts  E.  M.  F.  is  connected  to  a 
condenser  whose  capacity  is  one  half  microfarad. 
What  quantity  of  electricity  will  be  stored? 

Solution:  30  volts  X  .0000005  farads  =  .000015 
coulombs.  This  solution  could  also  be  given  direct- 
ly in  micro-quantities,  thus:  30  volts  X  %  micro- 
farad =  15  microcoulombs. 

A  condenser  is  charged  with  7.5  microcoulombs 
under  an  E.  M.  F.  of  15  volts.  What  is  its  capacity  ? 

Solution:  7.5  microcoulombs  -r-  15  volts  =  .5 
microfarad.  Ans. 

What  E.  M.  F.  is  required  to  charge  a  condenser 
whose  capacity  is  .1  microfarad  with  21  microcoul- 
ombs of  electricity? 

Solution:  21  microcoulombs  -f-  .1  microfarad  = 
210  volts.  Ans. 

By  connecting  condensers  in  parallel  the  resulting 
capacity  is  the  sum  of  their  individual  capacities. 
When  they  are  connected  in  series  the  resulting  capa- 
city equals  1  divided  by  the  sum  of  the  reciprocals 
of  their  individual  capacities.  It  will  be  noticed  that 
these  laws  of  condenser  connections  are  the  inverse 
of  those  for  the  parallel  and  series  connection  of  re- 
sistances. 


CONDENSERS.  123 

When  applying  a  direct  current  to  a  condenser, 
as  in  the  above  examples,  it  flows  until  the  increasing 
charge  opposes  an  E.  M.  F.  equal  to  that  of  the 
charging  current. 

With  an  alternating  current  a  charge  would  he 
surging  in  and  out  of  the  condenser,  so  that  a  real 
current  will  be  flowing  on  the  charging  wires  in 
spite  of  the  fact  that  the  actual  resistance  of  a  con- 
denser, in  ohms,  is  practically  infinite. 

Rule  91.  The  alternating:  current  in  a  circuit  hav- 
ing capacity  equals  the  product  of  6.2S32,  the  fre- 
quency, the  capacity,  and  the  applied  voltage. 

EXAMPLE. 

Find  the  current  produced  by  an  E.  M.  F.  of  50 
volts  and  a  frequency  of  60  cycles  per  second  in  a 
circuit  whose  capacity  is  125  microfarads. 

Solution:  The  capacity  125  microfarads  equals 
.000125  farads. 

6.2832  X  60  X  .000125  X  50  =  2.3562  amperes. 
Ans. 

Rule  92.  The  alternating:  E.  M.  F.  required  to  he 
impressed  upon  a  circuit  of  a  given  capacity  in 
order  to  produce  a  certain  current  is  equal  to  the 
current  divided  hy  6.2S32  times  the  product  ol  the 
capacity  and  the  frequency. 

EXAMPLE. 

Find  the  E.  M.  F.  necessary  to  produce  an  alter- 
nating current  of  50  amperes  at  50  cycles  per  sec- 
ond in  a  circuit  of  80  microfarads  capacity. 

Solution:  6.2832   X   .000080  X  50  =  .0251328 


124  ARITHMETIC  OF  ELECTRICITY. 

Dividing  the  current  50  amp.  by  .0251328  gives  2,000 
volts,  nearly. 

Since,  in  a  condenser  circuit,  a  real  current  flows 
under  a  given  E.  M.  F.,  the  circuit  may  be  treated  as 
though  it  was  of  a  resistance  such  as  would  allow 
the  given  current  to  flow. 

Rnle  93.  The  resistance  due  to  capacity  equals  1 
divided  by  the  product  of  6.2832,  the  frequency  and 
the  capacity. 

EXAMPLE. 

Find  the  capacity  resistance  of  a  circuit  having  a 
frequency  of  60  cycles  per  second  and  a  capacity  of 
50  microfarads. 

Solution:  6.2832  X  60  X  .000050  =  .01885. 
1  -T-  .01885  =  53  ohms,  very  nearly. 

By  comparing  this  Eule  93  for  capacity  resistance 
with  Eule  86  on  page  120,  which  is  for  inductive  re- 
sistance, it  will  be  seen  that  they  are  mutually  recip- 
rocal and  hence  the  effect  of  capacity  is  directly 
opposite  to  that  of  self-induction  and  vice  versa. 

It  follows  from  this  that  it  is  possible,  by  the 
proper  proportioning  of  the  inductance  and  the 
capacity,  to  have  their  effects  neutralized,  and  when 
this  adjustment  is  effected  the  current  will  be  con- 
trolled by  the  volts  and  ohmic  resistance  the  same 
as  if  it  were  a  direct  current  circuit. 

The  impedance  is  the  apparent  resistance  of  an 
alternating  current  circuit. 


CONDENSERS.  125 

Rule  94.  To  flnd  the  impedance  of  a  circuit  vrliose 
•r>liiiiic  resistance  can  be  neglected  and  ivhicrli  has 
an  inductance  and  a  capacity  in  series,  calculate 
both  the  inductive  resistance  and  the  capacity  re- 
sistance; their  difference  Trill  be  the  impedance. 

EXAMPLE. 

Find  the  current  produced  by  an  alternating  E. 
M.  F.  of  40  volts  on  a  circuit  of  slight  ohmic  resist- 
ance whose  capacity  is  100  microfarads,  the  frequency 
being  60  cycles  per  second,  and  having  in  series  an 
inductance  of  .02  henry. 

Solution:  Inductive  resistance  is  6.2832  X  60 
X  .02  =  7.42 ;  capacity  resistance  is  1  -=-  6.2832  X 
60  X  .000100  =  1  -^  .0377  =  26.52. 

Impedance  =  26.52  —  7.42  =  19.1  ohms. 

Current,  by  Ohm's  Law,  =  40  -f-  19.1  =  2.08  am- 
peres. Ans. 


CHAPTER  XIII. 

DEMONSTRATION   OF  RULES. 

In  the  following  chapter  we  give  the  demonstra- 
tion of  some  of  the  rules.  As  this  is  not  within  the 
more  practical  portion  of  the  work,  algebra  is  used 
in  some  of  the  calculations.  It  is  believed  that  rules 
not  included  in  this  chapter,  if  not  based  on  experi- 
ment, are  such  as  to  require  no  demonstration  here. 

Rule  i  to  6,  pages  13  and  14.  Ohm's  law  was 
determined  experimentally,  and  all  the  six  forms 
given  are  derived  by  algebraic  transposition  from 
the  first  form  which  is  the  one  most  generally  ex- 
pressed. 

Rule  8,  page  19.  This  is  simply  the  expression  of 
Ohm's  law  as  given  in  Rule  1,  because  in  the  case  of 
divided  circuits  branching  from  and  uniting  again 
at  common  points,  it  is  obvious  that  the  difference 
of  potential  is  the  same  for  all.  Hence  the  ratio  as 
stated  must  hold. 

Rule  9,  page  20.  This  rule  is  deduced  from  Rule 
8.  It  first  expresses  by  fractions  the  relations  of 
the  current.  Next  these  fractions  are  reduced  to  a 
common  denominator,  so  as  to  stand  to  each  other  in 


128  ARITHMETIC  OF  ELECTRICITY. 

the  ratio  of  their  numerators.  By  applying  the  new 
common  denominator  made  up  of  the  sum  of 
the  numerators  the  ratio  of  the  numerators 
is  unchanged,  and  the  ratio  of  the  new  fractions  is 
the  same  as  that  of  their  numerators,  while  by  this 
operation  the  sum  of  the  new  fractions  is  made 
equal  to  unity.  Thus  by  multiplying  the  total  cur- 
rent by  the  respective  fractions  it  is  divided  in  the 
ratio  of  their  numerators,  which  are  in  the  inverse 
ratio  of  the  resistances  of  the  branches  of  the  circuit 
and  as  the  sum  of  the  fractions  is  unity,  the  sum  of 
the  fractions  of  the  current  thus  deduced  is  equal  to 
the  original  current. 

Rule  10,  page  22.  Resistance  is  the  reciprocal  of 
conductance.  By  expressing  the  sum  of  the  recip- 
rocals of  the  resistances  of  parallel  circuits  we  ex- 
press the  conductance  of  all  together.  The  recipro- 
cal of  this  conductance  gives  the  united  resistance. 

Rule  11,  page  22.  This  is  a  form  of  Rule  10. 
Call  the  two  resistances  x  and  y.  The  sum  of  their 
reciprocals  is  *  +  *  which  is  the  conductance  of  the 
two  parallel  circuits  or  parts  of  circuits.  Reducing 
them  to  a  common  denominator  we  have:  -£.  +  .*. 
which  equals  ~~,  whose  reciprocal  is 


Rule  17,  page  31.  Taking  the  diameter  of  a  wire 
as  d,  its  cross  sectional  area  is  —-.  The  resistance 
is  inversely  proportional  to  this  or  varies  directly 
with  =^.  As  the  resistance  of  a  conductor 


DEMONS  TEA  TION  OF  E  ULES.  129 

Taries  also  with  its  length  and  specific  resistance  we 
have  as  the  expression  for  resistance: 

Sp.  Bes.  X  1.2737  X  1 
da 

Rule  18,  page  32.  Assume  two  wires  whose 
lengths  are  I  and  Z15  their  cross  sectional  areas  a  and 
#!,  their  specific  resistances  s  and  51?  and  their  resis- 
tances r  and  r^.  From  preceding  rules  we  have  for 
each  wire:  r  =  s  ~  (1)  and  n  =  s,  ^  (2). 
Dividing  (1)  by  (2)  we  have: 


If  we  take  the  reciprocal  of  either  member  of  this 
equation  and  multiply  the  other  member  thereby  it 
will  reduce  it  to  unity,  or: 

-*v  —  v—  v^-1 

r  x   *x  *  it  x  a  .  - 

For  convenience  this  is  put  into  a  shape  adapted 
for  cancellation. 

Rule  20,  page  38.  This  is  merely  the  expression 
of  Ohm's  Law,  Eule  3. 

Rule  22,  page  40.  Call  the  drop  e,  the  combined 
resistance  of  the  lamps  E,  and  the  resistance  of  the 
leads  x.  Then  as  the  whole  resistance  is  expressed 
as  1  00  (because  the  work  is  by  percentage)  the  differ- 
ence of  potential  for  the  lamps  is  100  —  e.  By 
Ohm's  law  we  have  the  proportion:  100-0  :  e  ::  R  : 
x  or 

eR 

x  ~  lOO^e 

Rule  25,  page  44.     From  Eule  22  we  have: 


130  ARITHMETIC  OF  ELECTRICITY. 

Call  the  resistance  of  a  single  lamp  r,  then  we 
have  by  Rule  12: 

E=S  (2) 

Substituting  this  value  of  R  in  equation  (1)  we 
have: 

—          er 

~  wx(lOO— e)  (3) 

From  Rule  24  we  have,  calling  the  cross-section  a: 
i  x  10.79 
~~F~  (4) 

Substituting  for  x  its  value  from  equation  (3)  we 
have: 

I  X  10.79  X  n  X  (100— e) 
a  =  J /K\ 

(5) 

But  as  I  expresses  the  length  of  a  pair  of  leads, 
not  the  total  length  of  lead  but  only  one-half  the 
total,  the  area  should  be  twice  as  great.  This  is 
effected  by  using  the  constant  10.79  X  2  =  21.58  in 
the  equation  giving: 

I  X  21.58  X  n  X  (100-e) 
er 

Rule  28,  page  48.  Assuming  the  converter  to  work 
with  100$  efficiency  (which  is  never  the  case),  the 
watts  in  the  primary  and  secondary  must  be  equal 
to  each  other  or: 

C2  R  =  (V  Rx,  and  R  =  ^  Rx, 

or  the  resistances  of  primary  and  secondary  are  in 
the  ratio  of  the  squares  of  the  currents.  The  direct 
ratio  is  expressed  by  the  ratio  of  conversion,  when 
squared  it  gives  the  ratio  of  the  squares  as  required. 


DEMONSTRATION  OF  RULES.  131 

Rule  37,  page  59.  Let  d  =  diameter  of  the  wire  in 
centimeters.  The  resistance  of  one  centimeter  of 
such  a  wire  in  ohms  =  Sp.  Resist.  X  10*6  X  ^.  The 
specific  resistance  is  here  assumed  to  be  taken  in 
microhms.  The  quantity  of  heat  in  joules  devel- 
oped in  such  a  wire  in  one  second  is  equal  to  the 
square  of  the  current  in  C.  G.  S.  units,  multiplied 
by  the  resistance  in  C.  G.  S.  units  and  divided  by 
4.16  X  107,  the  latter  division  effecting  the  reduc- 
tion to  joules.  1  ohm  =  109  C.  G.  S.  units  of  re- 
sistance. Multiplying  the  expression  for  ohmic 
resistance  by  109  we  have:  Sp.  Resist.  X  10*  X  ~_. 
1  ampere  =  lOi  0.  G.  S.  unit.  If  we  express  the 
current  in  amperes  we  must  multiply  it  by  10"1,  in 
other  words  take  one-tenth  of  it.  Our  expression 
then  becomes  for  heat  developed  in  one  second 

/JL.Y  x- Sp-  Resist-  x  iQ3  x  4 

V 10  /    "     IT  da  X  4.16  X  10* 

The  area  of  one  centimeter  of  the  wire  is  *d 
square  centimeters.  The  heat  developed  per  square 
centimeter  is  found  by  dividing  the  above  expression 
by  ird  giving: 

sjz_  \  *  ySp.  Resist.  X  103  X  4 
\10)  ir'  d3  X  4.16  X  107 

The  heat  developed  is  opposed  by  the  heat  lost 
which  we  take  as  equal  to  njW  per  square  centimeter 
per  degree  Cent,  of  excess  above  surrounding  me- 
dium. Therefore  taking  t°  as  the  given  tempera* 


132  ARITHMETIC  OF  ELECTEICITY. 

ture  cent,  we   may  equate  the  loss  with  the  gain 
thus: 


t°      _  /_£  \a 
4000    ~   \10j 


x  Sp  Resist.  X  103  X  4 
'        IT*  d3  X  4.16  X  107 


_  c8  X  Sp.  Resist.  X  103  X  4  x  40000  _ 
**  x  4.16  X  10'  X  t° 

ca  X  Sp.  Resist,  x  .00039 

t° 

• 

Rule  52,  page  69.     Call  the  external  resistance  r; 
number  of  cells  n;  resistance  of  one  cell  R;  E.  M. 
F.  of  one  cell  E;  E.  M.  F.  of  outer  circuit  e. 
Then  from  Ohm's  law  we  have: 

nE 
~  nR  +  r  (1) 

which  reduces  to: 

n=!E=QR  (2) 

but  C  r  =  e. 


•';  n  =  W=cJi  (3) 

Rule  54,  page  72.  This  rule  is  deduced  from  the 
following  considerations.  The  current  being  con- 
stant the  work  expended  in  the  battery  and  external 
circuit  respectively  will  be  in  proportion  to  their 
differences  of  potential  or  E.  M.  F's.  But  these 
are  proportional  to  the  resistances.  Therefore  the 
resistance  of  the  external  circuit  r  should  be  to  the 
resistance  of  the  battery  R  as  efficiency:  1  —  effi- 
ciency or  r  :  R  ::  efficiency  :  1  —  efficiency  or 

r-       The  rest  °f  th 


from  Ohm's  law. 


DEMONSTRATION  OF  EULES.  133 

Rule  5  7,  page  74.  This  rule  gives  the  nearest  ap- 
proximation attainable  without  irregular  arrange- 
ment of  cells.  By  placing  some  cells  in  single  series 
and  others  two  or  more  in  parallel,  an  almost  exact 
arrangement  for  any  desired  efficiency  can  be  ob- 
tained. Such  arrangement  are  so  unusual  that  it  is 
not  worth  while  to  deduce  any  special  rule  for 
them.  Thus  taking  the  example  given  on  page 
74  the  impossible  arrangement  of  1.4  cells  in 
parallel  and  63  in  series  would  give  the  desired 
current  and  efficiency.  The  same  result  can 
be  obtained  by  taking  72  cells  in  36  pairs  with  a  re- 
sistance of  36  x  &  =  3  ohms,  and  adding  to  them 
27  cells  in  series  with  a  resistance  of  27  X  £  =  4^ 
ohms,  a  total  of  7l  ohms.  The  E.  M.  P.  is  equal  to 
(36  +  27)  X  2  =  126  volts.  The  total  cells  are 
72  +  27  =  99. 

Rule  58,  page  76.  Ono  coulomb  of  electricity  lib- 
erates from  an  electrolyte  .000010384  gram  of 
hydrogen.  This  has  been  determined  experimen- 
tally. Let  H  be  the  heat  liberated  by  the  chemical 
combining  weight  of  any  body  combining  with 
another.  H  is  taken  in  kilogram  calories.  Hence  it 
follows  that  for  a  quantity  of  the  substance  equal  to 
.000010384  gram  X  chemical  combining  weight,  the 
heat  liberated  will  be  equal  to  H  X  .000010384, 
which  corresponds  to  a  number  of  kilogram  meters 
of  work  expressed  by  .000010384  X  H  X  424.  The 
work  done  by  a  current  in  kilogram-meters  = 


134  ARITHMETIC  OF  ELECTRICITY. 


volts  X^oulombs  OJ.  f()r 

presses  the  work  done  by   one  coulomb.     Let  the 
volts  =  E,  and  equate  these  two  expressions: 
ofi  =  .000010384  X  H  X  424, 
which  reduces  to 

E  =  H  X  .043. 

Rule  6i,  page  78.  For  the  work  (in  kilogram- 
meters)  done  by  a  current  (volt-coulombs)  we  have 
the  general  expression: 

_       volts  x  coulombs  _  „  E  Q 

~W~     -     °r  9^  (1) 

Making  W  =  1  (i.  e.  one  kilogram-meter)  and 
transforming,  we  have,  as  the  coulombs  correspond- 
ing to  1  kilogram-meter: 

_        9.81 

«•  -r  (3) 

One  coulomb  of  electricity  liberates  a  weight  (in 
grams)  of  an  element  equal  to  the  product  of  the 
following:  .000010384  X  equivalent  of  element  in 
question  X  number  of  equivalents  -*-  valency  of  the 
element.  Therefore,  the  coulombs  corresponding 
to  one  kilogram-meter,  liberates  this  weight  multi- 
plied by  ~  or,  indicating  weight  by  G-, 


G  —  -Q000103^  X  equiv.  X  number  equiv.  x  9.81 

valency  E  (3) 

but  .000010384  X  9.81  =  .000101867. 

_  _  equiv.  X  n  X  .000101867 
•  *  E  X  valency  (4) 

Rule  73,  page  98-99.  The  voltage  of  an  armature  of 


DEMONSTRATION  OF  RULES.  135 

a  definite  number  of  turns  of  wire  and  a  fixed  speed, 
varies  with  the  lines  included  within  its  longitu- 
dinal area,  as  such  lines  are  cut  in  every  revolution. 
These  lines  vary  with  its  area,  and  the  latter  varies 
with  the  square  of  its  linear  dimensions. 

To  maintain  a  constant  voltage  if  the  size  is 
changed,  the  number  of  turns  must  be  varied  in- 
versely as  the  square  of  the  linear  dimensions. 
This  ensures  the  cutting  of  the  same  number  of 
lines  of  force  per  revolution. 

If,  therefore,  its  size  is  reduced  from  x  to  ^  the 
turns  of  wire  must  be  changed  from  x  to  oft.  The 
relative  diameters  of  the  two  sizes  of  wire  is  found 
by  dividing  a  similar  linear  dimension  by  the  rela- 
tive size  of  the  wire.  But  |  •*•  x*  =  ^  =  diameter  of 
the  wire  for  maintenance  of  a  constant  voltage  with 
change  of  size. 

The  capacity  of  a  wire  varies  with  the  square  of 
its  diameter  and  (-^)2  =  ^. 

Therefore  the  amperage,  if  a  constant  voltage  is 
maintained,  will  vary  inversely  as  the  sixth  power  of 
the  linear  dimensions  of  an  armature. 


CHAPTER  XIV. 

NOTATION  IN"   POWEKS   OF  TEN. 

THIS  adjunct  to  calculations  has  become  almost 
indispensable  in  working  with  units  of  the  C.  G.  S. 
system.  It  consists  in  using  some  power  of  10  as  a 
multiplier  which  may  be  called  the  factor.  The 
number  multiplied  may  be  called  the  characteristic. 
The  following  are  the  general  principles. 

The  power  of  10  is  shown  by  an  exponent  which 
indicates  the  number  of  ciphers  in  the  multiplier. 
Thus  102  indicates  100;  108  indicates  1000  and  so 
on. 

The  exponent,  if  positive,  denotes  an  integral 
number,  as  shown  in  the  preceding  paragraph.  The 
exponent,  if  negative,  denotes  the  reciprocal  of  the 
indicated  power  of  10.  Thus  10~2  indicates  TST>',  10'3 
indicates  rrsW  and  so  on. 

The  compound  numbers  based  on  these  are  re- 
duced by  multiplication  or  division  to  simple  expres- 
sions. Thus:  3.14  X  107  =  3.14  X  10,000,000  = 
31,400,000.  3.14X10-^1^^or1-^^.  Re- 

gard  must  be  paid  to  the  decimal  point  as  is  done 
here. 


NOTATION  IN  POWERS  OF  TEN.  137 

To  add  two  or  more  expressions  in  this  notation 
if  the  exponents  of  the  factors  are  alike  in  all  re- 
spects, add  the  characteristics  and  preserve  the  same 
factor.  Thus: 

(51  X  106)  +  (54  X  106  )=  105  X  10». 

(9.1  X  10-9)  +  (8.7  X  10-9)  =  17.8  X  10-'. 

To  subtract  one  such  expression  from  another, 
subtract  the  characteristics  and  preserve  the  same 
factor.  Thus: 

(54  X  106)  -  (51  X  106)  =  3  X  106. 

If  the  factors  have  different  exponents  of  the 
same  sign  the  factor  or  factors  of  larger  exponent 
must  be  reduced  to  the  smaller  exponent,  by  factor- 
ing. The  characteristic  of  the  expression  thus 
treated  is  multiplied  by  the  odd  factor.  This  gives 
a  new  expression  whose  characteristic  is  added 
to  the  other,  and  the  factor  of  smaller  exponent  is 
preserved  for  both. 

Thus: 

(5  X  107)  +  (5  X  109)  =  (5  X  107)  +  (5  X  100  X 
107)  =  505  X  107. 

The  same  applies  to  subtraction.     Thus: 

(5  X  109)  -  (5  x  107)  =  (5  X  100  X  107)  -  (5  X 
107)  =  495  X  107. 

If  the  factors  differ  in  sign,  it  is  generally  best  to 
leave  the  addition  or  subtraction  to  be  simply  ex- 


138  ARITHMETIC  OF  ELECTRICITY. 

pressed.     However  by  following  the  above  rule  it 
can  be  done.     Thus: 

Add  5  X  10'2  and  5  X  103. 

5  X  10s  =  5  X  105  X  10-2  :  (5  X  105  X  lO'2)  +  (5  X 
lO'2)  =  500005  X  lO'2.  This  may  be  reduced  to  a 
fraction  ^f  =  5000.05. 

To  multiply  add  the  exponents  of  the  factors,  for 
the  new  factor,  and  multiply  the  characteristics  for 
a  new  characteristic.  The  exponents  must  be  added 
algebraically:  that  is,  if  of  different  signs  the  numer- 
ically smaller  one  is  subtracted  from  the  other  one, 
its  sign  is  given  the  new  exponent. 

Thus: 

(25  X  106)  X  (9  X  108)  =  225  X  1014. 
(29  X  10-8)  X  (11  X  107)  =  319  X  10-1. 
(9  X  108)  X  (98  X  10-2)  -  882  X  106. 

To  divide,  subtract  (algebraically)  the  exponent 
of  the  divisor  from  that  of  the  dividend  for  the  ex- 
ponent of  the  new  factor,  and  divide  the  character- 
istics one  by  the  other  for  the  new  characteristic. 
Algebraic  subtraction  is  effected  by  changing  the 
sign  of  the  subtrahend,  subtracting  the  numer- 
ically smaller  number  from  the  larger,  and  giving 
the  result  the  sign  of  the  larger  number.  (Thus  to 
subtract  7  from  5  proceed  thus:  5  —  7  =  —2.) 

Thus: 

(25  X  106)  •*-  (5  X  108)  =  5  X  10-2 
(28  X  10-8)  •*•  (5  X  108)  =  5.6  X  10'u. 


TABLES. 


| 

! 


140 
II.— EQUIVALENTS   OF   UNITS   OF   AREA. 


Square 
Millimeter 

Square 
Centimet'r 

Circular 
Mil. 

Square 
Mil. 

Square 
Inch. 

Square 
Foot. 

.0000108 

Square  Millimeter 

1 

0.01 

1973.6 

1550.1 

.00155 

Square  Centimeter 

100 

1 

197,361 

155,007 

.155007 

.C01076 

Circular  Mil. 

.000507 

.0000051 

1 

.78540 

8X10-' 

Square  Mil. 

.000645 

.0000065 

1.2733 

1 

.000001 

Square  Inch 

645.132 

6.451 

1,273,238 

1,000,000 

1 

.006944 

Square  Foot 

^,898.9 

928.989 

144 

1 

HI.— EQUIVALENTS  OF  UNITS  OF  VOLUME. 


Cubic 
Inch 

Fluid 
Ounce 

Gallon 

Cubic 
Foot 

Cubic 
Yard 

Cu.  Cen- 
timeter 

Liter 

Cubic 
Meter 

Cubic  Inch 

1 

.554112 

.004329 

.000578 

16.3862 

.016386 

Fluid  Oz. 

1.80469 

1 

.007812 

.001044 

29.5720 

.029572 

Gallon 

231 

128 

1 

.133681 

.00495 

8785.21 

3.78521 

.003785 

Cubic  Ft. 

1728 

957.506 

7.48052 

1 

.037037 

28315.3 

28.3153 

.028315 

Cubic  Yd. 

46,656 

25,852.6 

201.974 

27 

1 

764,505 

764.505 

.764505 

Cu.  Centi. 

.061027 

.033816 

.000264 

.000035 

1 

.001 

.000001 

Liter 

61.027 

33.8160 

.264189 

.035317 

1000 

1 

.001 

Cu.  Meter 

61027 

33816 

264.189 

35.3169 

1.3080 

1000 

1 

141 
IV.— EQUIVALENTS   OF   UNITS   OF  WEIGHT. 


Grain 

Grain. 

Troy 
Ounce. 

Pound 
Avs. 

Ton. 

Milli- 
gram. 

Gram. 

Kilo- 
gram. 

Metric 
Ton. 

1 

.020833 

.000143 

64.799 

.064799 

.000065 

TroyOunce 

480 

1 

.068641 

31,103.5 

31.1035 

.031104 

PoundAvs. 

7,000 

14.5833 

1 

.000447 

453.593 

.453593 

.000454 

Ton 

32,666.6 

2240 

1 

.001016 

1.01605 

Milligram 

.015432 

.000032 

.000002 

1 

.001 

.000001 

Gram 

15.4323 

.032151 

.002205 

1000 

1 

.001 

Kilogram 

15,432.3 

32.1507 

2.20462 

.000984 

1,000,000 

1000 

1 

.001 

Metric  Ton 

32,150.7 

2204.62 

.98421 

1,000,000 

1000 

1 

142 


V.-EQUIVALENT8  OP  UNITS 


Erg. 

Meg- 
erg. 

Gram-de- 
gree C. 

Kilogram- 
degree  C. 

Pound- 
degree  C. 

Pound- 
degree  F. 

Erf. 

1 

.000001 

Meg.-erg. 

1,000,000 

1 

.024068 

.000024 

.000053 

.000095 

Gram-degree  C. 

41.5487 

1 

.001 

.002205 

.003968 

Kilogram-degreeC. 

41,548.T 

1000 

1 

2.2046 

8.9683 

Pound-degree  C. 

18,846.5 

453.59 

.45359 

1 

1.8 

Pound-degree  F. 

10,470.1 

251.995 

.251995 

.555556 

•v 

Watt-Second. 

107 

10 

.24068 

.000241 

.000531 

.000955 

Gram-centimeter. 

981 

.000981 

.0000235 

Kilogram-meter. 

98.1X108 

98.1 

2.86108 

.002861 

.005205 

.009370 

Foot-Pound. 

18.5626 

.326425 

.000826 

.000720 

.001295 

Horse-Power-Sec. 

English. 

7459.43 

179.486 

.179486 

.8957 

.71243 

Horse-Power-Sec. 
Metric. 

7357.5 

177.075 

177.075 

.890375 

.70275 

143 


OF  ENERGY  AND  WORK. 


Watt- 
Second. 

Gram- 
Centim'tr. 

Kilogram- 
meter. 

Foot- 
Pound. 

Horse- 
power- 
second 
English. 

Horse- 
power- 
second 
Metric. 

10-T 

.001019 

Erg. 

.1 

1019.87 

.010194 

.078784 

.000184 

.000186 

Meg-erg. 

4.15487 

42,858.5 

.428585 

3.06355 

.00557 

.005647 

Gram-degree  C. 

4154.87 

428.535 

8068.55 

5.57 

5.64703 

Kilogram-degree  C. 

1834.65 

192.114 

1889.6 

2.52653 

2.56149 

Pound-degree  C. 

1047.08 

106.730 

772 

1.40364 

1.42805 

Pound-degree  F. 

1 

10,193.7 

.101937 

.737387 

.0018406 

.0018592 

Watt-Second. 

.000098 

1 

.00001 

.000072 

Gram-Centimeter. 

9.81 

100,000 

1 

7.23828 

.018152 

.018334 

Kilogram-meter. 

1.35626 

13,825.8 

.188258 

1 

.0018182 

.001843 

Foot-Pound. 

745.943 

76.0392 

550 

1 

1.01888 

Horse-Power-Sec. 
English. 

735.75 

75 

542.496 

.986356 

1 

Horse-Power-Sec. 
Metric. 

144 

VI.— TABLE    OF    SPECIFIC    RESISTANCES    IN    MICROHMS     AND 
OF  COEFFICIENTS   OF   SPECIFIC  RESISTANCES  OF  METALS. 


Specific 

Coeffi- 

Specific 

Coeffi- 

Resist- 
ance. Mi- 

cients of 
Sp.  Res. 

Resist- 
ance. Mi- 

cients of 
Bp.  Res. 

crohms. 

crohms. 

Annealed  Silver  

1.521 

.9412 

Annealed  Nickel  .... 

12.60 

7.7970 

Hard  Silver 

1.652 
1.616 

.0223 
.0000 

Compres'd  Tin  
Lead  .... 

18.36 
19.85 

8.2673 
12.2834 

Annealed  Copper.... 

Hard  Copper  

1.652 

.0223 

"          Antimony 

85.90 

22.2153 

Annealed  Gold  

2.081 

.2877 

"         Bismuth  . 

182.70 

82.1170 

Hard  Gold     

2.118 

.8107 

Liquid  Mercury  

99.74 

61.7203 

Annealed  Aluminum.. 

2.945 

1.8224 

2  Silver,  1  Platinum. 

24.66 

15.2599 

Compressed  Zinc  .... 

5  689 

8.5204 

German  Silver  .... 

21  17 

18  1002 

Annealed  Platinum  .  . 

9.158 

5.6671 

2  Gold,  1  Silver  .... 

10.99 

6.8008 

"        Iron  

9.825 

6.0798 

SPECIFIC  RESISTANCE  OF  SOLUTIONS  AND  LIQUIDS. 

MATTHIESSEN  AND  OTHERS. 


Names  of  Solutions. 

Temper- 
ature 
Centi- 
grade. 

Temper- 
ature 
Fahren- 
heit. 

Specific 
Resistance. 
Ohms. 

Copper  Sulphate,  concentrated 

9° 

48.2° 

29.82 

"                with  an  equal  volume  of  water 

it 

it 

46.54 

it 

with  three  volumes  of  water 

it 

it 

77.68 

Common  Salt, 

concentrated  

13° 

554° 

5.98 

it 

with  an  equal  volume  of  water  .  . 

1C 

'< 

6.00 

" 

with  two  volumes  of  water.... 

II 

it 

9.24 

" 

with  three  volumes  of  water.  . 

II 

•• 

11.89 

Zinc  Sulphate, 

concentrated  

14° 

57.2° 

28.00 

it 

with  an  equal  volume  of  water.  . 

'< 

22.75 

" 

with  two  volumes  of  water  

it 

it 

29.75 

Sulphuric  Acid 

,  concentrated  

14.3° 

57.8° 

5.32 

50.5#,  Specific  Gravity  1.393.  .  . 

14.5° 

58.1° 

1.086 

" 

29.6*.  Specific  Gravity  1.215.  .  . 

12.3° 

54.5° 

.83 

tt 

12*  Specific    Gravity  1.080... 

12.8° 

55.0° 

1.368 

Nitric  Acid,  Specific  Gravity  1.36  (Blavier)  .... 

14° 

57.2° 

1.45 

H 

(i            •(         ti 

24° 

75.2° 

1.22 

Distilled  Water 

,  (Temp'ture  unknown)  (Pouillet) 

988. 

VIL— RELATIVE    RESISTANCE    AND    CONDUCTANCE    OF    PURE 
COPPEE  AT  DIFFERENT  TEMPERATURES. 

KATTHTESSEX. 


$4 

go 

§3* 

!l 

Relative 
Resistance. 

Relative 
Conductance 

II 

1 

Relative 
Resistance. 

Relative 
Conductance 

0' 

82' 

1. 

1. 

16° 

60.8« 

1.06168 

.9419 

i 

83.8 

1.00881 

.99620 

17 

62.6 

1.06568 

.93841 

2 

85.8 

1.00756 

.9925 

15 

64.4 

1.06959 

.93494 

8 

87.4 

1.01185 

.98878 

19 

66.2 

1.07356 

.93148 

4 

89.2 

1.01515 

.98508 

20 

68. 

1.07754 

.92804 

5 

41 

1.01896 

.98189 

21 

69.8 

1.08152 

.92462 

6 

42.8 

1.0228 

.97771 

22 

71.6 

1.08558 

.92120 

7 

44.6 

1.02663 

.97406 

23 

78.4 

1.08954 

.91782 

8 

46.4 

1.03048 

.97042 

24 

75.2 

1.09356 

.91445 

9 

48.2 

1.08485 

.96679 

25 

77. 

1.09759 

.9111 

10 

50 

1.08822 

.96319 

26 

78.8 

1.10162 

.90776 

11 

61.8 

1.04210 

.95960 

27 

80.6 

1.10567 

.90443 

12 

58.6 

1.04599 

.95603 

28 

82.4 

1.10972 

.90113 

18 

55.4 

1.0499 

.95247 

29 

84.2 

1.11882 

.89784 

14 

57.2 

1.05381 

.94893 

36 

86. 

1.11785 

.89457 

15 

59 

1.05774 

.94541 

146 

VIII.— AMERICAN    WIBE    GAUGE    TABLE. 

Properties  of  Copper  Wire  :   Specific  Gravity,  8.8T8  ;   Specific  Conductivity,  1.T66  at  75°,  F. 


I 

SIZE. 

WEIGHT  AND  LENGTH. 

RESISTANCE. 

i'i| 

I  Gauge  Num 

III! 

°lll 

Diam- 
eter in 
Mils. 

Square  of 
Diameter 
or  circular 
Mils. 

Grains 
per 
Foot. 

Po'nds 
per 
1000 
Feet. 

Feet 
per 
Pound. 

Ohms 
per  1000 
Feet. 

Feet 
Ohm 

Ohms  per 
Pound. 

0000 

460.000 

211600.0 

4477.2 

639.60 

1.564 

.051 

19929.7 

.0000785 

480 

000 

409.640 

167804.9 

3550.5 

607.22 

1.971 

.063 

15804.9 

.000125 

262 

00 

864.800 

138079.0 

2815.8 

402.25 

2.486 

.080 

12534.2 

.000198 

208 

0 

824.950 

105592.5 

2236.2 

819.17 

8.183 

.101 

9945.8 

.000815 

165 

1 

289.300 

83694.49 

1770.9 

252.98 

8.952 

.127 

7882.8 

.000501 

180 

2 

257.630 

66873.22 

1404.4 

200.63 

4.994 

.160 

6251.4 

.000799 

108 

8 

229.420 

52688.58 

1118.6 

159.09 

6.285 

.202 

4957.8 

.001268 

81 

4 

204.810 

41742.57 

888.2 

126.17 

7.925 

.254 

8931.6 

.002016 

65 

5 

181.940 

88102.16 

700.4 

100.05 

9.995 

.821 

8117.8 

.008206 

62 

6 

162.020 

26250.48 

555.4 

79.84 

12.604 

.404 

2472.4 

.005098 

41 

7 

144.280 

20816.72 

440.4 

62.92 

15.898 

.609 

1960.6 

.008106 

82 

8 

128.490 

16509.68 

849.8 

49.90 

20.040 

.648 

1555.0 

.01289 

26 

9 

114.480 

13094.22 

277.1 

89.58 

25.265 

.811 

1288.8 

.02048 

20 

10 

101.890 

10381.57 

219.7 

81.88 

81.867 

1.028 

977.8 

.03259 

16 

11 

90.742 

8234.11 

174.2 

24.89 

40.176 

1.289 

775.5 

.05181 

18 

12 

80.808 

6529.93 

188.2 

19.74 

50.659 

1.626 

615.02 

.08287 

10.2 

18 

71.961 

6178.89 

109.6 

15.65 

68.898 

2.048 

488.25 

.18087 

8.1 

14 

64.084 

4106.75 

86.87 

12.41 

80.580 

2.585 

886.80 

.20880 

6.4 

15 

57.068 

8256.76 

68.88 

9.84 

101.626 

8.177 

806.74 

.83183 

5.1 

16 

50.820 

2582.67 

64.67 

7.81 

128.041 

4.582 

248.25 

.52638 

4.0 

17 

45.257 

2048.19 

48.88 

6.19 

161.551 

5.188 

192.91 

.83744 

8.2 

18 

40.308 

1624.38 

84.87 

4.91 

208.666 

6.536 

152.99 

1.8312 

2.5 

19 

85.390 

1252.45 

26.50 

8.786 

264.186 

8.477 

117.96 

2  2392 

1.96 

20 

81.961 

1021.51 

21.60 

8.086 

824.045 

10.394 

96.21 

8.3488 

1.60 

21 

28.462 

810.09 

17.14 

2.448 

408.497 

18.106 

76.80 

5.8539 

1.28 

22 

25.347 

642.47 

13.59 

1.942 

514.938 

16.525 

60.51 

8.5099 

1.08 

23 

22.571 

509.45 

10.77 

1.589 

649.773 

20.842 

47.98 

18.884 

.80 

24 

20.100 

404.01 

8.55 

1.221 

819.001 

26.284 

88.05 

21.524 

.63 

25 

17.900 

820.41 

6.77 

.967 

1084.126 

88.185 

80.18 

84.298 

.60 

26 

15.940 

254.06 

5.88 

.768 

1302.088 

41.789 

28.93 

54.410 

.40 

27 

14.195 

201.49 

4.26 

.608 

1644.737 

52.687 

18.98 

86.657 

.81 

28 

12.641 

159.79 

8.89 

.484 

2066.116 

66.445 

15.05 

137.283 

.25 

29 

11.257 

126.72 

2.69 

.884 

2604.167 

83.752 

11.94  , 

218.104 

.20 

80 

10.025 

100.50 

2.11 

.802 

8311.258 

105.641 

9.466 

849.805 

.16 

81 

8.928 

79.71 

1.67 

.289 

4184.100 

188.191 

7.508 

557.286 

.13 

82 

7.950 

63.20 

1.83 

.190 

5268.158 

168.011 

5.952 

884.267 

.098 

83 

7.080 

50.18 

1.06 

.151 

6622.517 

211.820 

4.721 

1402.78 

.078 

84 

6.804 

89.74 

.847 

.121 

8264.468 

267.165 

8.743 

2207.98 

.062 

85 

5.614 

81.52 

.658 

.094 

10688.30 

386.81 

2.969 

3583.12 

.049 

86 

6.000 

25.00 

.525 

.075 

18333.33 

424.65 

2.355 

5661.71 

.039 

87 

4.453 

19.88 

.420 

.060 

16666.66 

535.88 

1.868 

8922.20 

.081 

88 

8.965 

15.72 

.815 

.045 

22222.22 

675.22 

1.481 

15000.5 

.025 

'    89 

8.531 

12.47 

.266 

.088 

26315.79 

851.789 

1.174 

22415.5 

.020 

40 

8.144 

9.88 

.210 

.080 

8883388 

1074.11 

.981 

85803.8 

.015 

147 

IX.— CHEMICAL  AND  THEEMO-CHEMICAL  EQUIVALENTS. 
FOBMATIOIT  or  OXIDES. 


Name  of  Compound. 

Formula. 

Valency. 

Chemical 
Equiv- 
alents. 

Combin- 
ing 
Weights. 

Thermo- 
Chemical 
Equiv- 
alents. 

Water    

H«O 

II 

18 

9 

84.5 

Iron  Protoxide      

FeO 

II 

72 

86 

84.5 

Iron  Sesquioxide  .  . 

Fe»O3 

III 

160 

53  3 

81.9x8 

Zinc  Oxide 

Zn  O 

11 

81 

40  5 

48  2 

Copper  Oxide    

CuO 

II 

79.4 

89.7 

19.2 

Mercury  Oxide 

HgO 

II 

218 

108 

15.5 

FOBMATION  or  SALTS. 


Name  of 

Va- 

Nitrates 

Sul- 

Chlo- 

Cya- 

Base. 

lency. 

phates 

rides 

nides. 

FOBMTJLA. 

Fe  (N03)2 

FeSO* 

FeCia 

FeCy  » 

Chemical  Equivalents  

180 

186 

127 

112 

Combining  Weights  
Thermo-Chemical  Equiv'lts 

90 
13.9 

68 
12.5 

68.5 
00 

66 
8.2 

FOBMTTLA 

Zn(NO3)2 

ZnSO± 

ZnC12 

ZnCy» 

Chemical  Equivalents  

189 

161 

136 

117 

Combining  Weights  
Thermo-Chemical  Equiv'lts 

94.5 
9.8 

80.5 
11.7 

68 
66.4 

58.5 
7.8 

FOBMTTLA. 

Cu(NO3)2 

CuSO* 

CuCl* 

CuCy* 

Copper 

II 

Chemical  Equivalents   
Combining  Weights  

187.4 
93.7 

159.4 
79.7 

134.4 
67.2 

125.4 
62.7 

Thermo-Chemical  Equiv'lts 

7.5 

9.2 

81.8 

7.8 

FOBMTTLA. 

Hg(N03)2 

HgSO* 

HgC12 

HgCya 

Mercury 

II 

Chemical  Equivalents  
Combining  Weights  
Thermo-Chemical  Equiv'lts 

324 
162 
7.5 

280 
140 
9.2 

271 
135.5 
9.45 

252 
126 
15.5 

148 
X.— CHEMICAL  AND  ELECTEO-CHEMICAL  EQUIVALENTS. 


Name 

Symbols 

Valen- 
cies 

Chemical 
Equivalents 

Combining 
Weights 

Electro- 
Chemical 
Equivalents 

Hydrogen 

H 

I 

1 

1 

.0105 

Gold 

Au 

III 

196.6 

65.5 

.6877 

Silver  

Ag 

I 

108 

108 

1.134 

Copper  (Cupric)  

Cu,, 

II 

63 

81.5 

.8307 

Mercury  (Mercuric)  .  .  . 
"      (Mercurous).. 
Iron(ferrlc)  
"    (ferrous) 

Hg,, 
Hg. 
Fe,,, 
Fe, 

n 
i 
in 
ii 

200 
200 
56 
56 

100 
200 

18.7 
28 

1.05 
2.10 
.1964 
.294 

Nickel  

Ni 

ii 

59 

29.5 

.8098 

Zinc 

Zn 

ii 

65 

82  5 

.3418 

Lead          

Pb 

ii 

207 

108.5 

1.0868 

Oxygen  

0 

ii 

16 

8 

084 

Chlorine  

Cl 

i 

85.5 

35.5 

.8728 

XI.— MAGNETIZATION  AND  MAGNETIC  TRACTION. 


B 

Lines  per 
eq.  cm. 

B,, 
Lines  per 
sq.  in. 

Dynes 
per 
sq.  centim. 

Grammes 
per 
sq.  centim. 

Kilogrs. 
per 
eq.  centim. 

Pounds 
per 
sq.  inch. 

1,000 

6,450 

89,790 

40.56 

.0456 

.577 

2,000 

12,900 

159,200 

162.3 

.1628 

2.808 

8,000 

19,850 

858,100 

865.1 

.8651 

5.190 

4,000 

25,800 

636,600 

648.9 

.6489 

9.228 

5,000 

82,250 

994,700 

1,014 

1.014 

14.89 

6,000 

88,700 

1,482,000 

1,460 

1.460 

20.75 

7,000 

45,150 

1,950,000 

1,987 

1.987 

28.26 

8,000 

51,600 

2,547,000 

2,596 

2.596 

86.95 

9,000 

58,050 

8,223,000 

8,286 

8.2S6 

46.72 

10,000 

64,500 

8,979,000 

4,056 

4.056 

57.68 

11,000 
12,000 
18,000 

70,950 
77,400 

88,850 

4,815,000 
5,730,000 
6,725,000 

4,907 
5,841 
6,855 

4.907 
5.841 
6.855 

69.77 
88.07 
97.47 

14,000 

90,300 

7,800,000 

7,550 

7.550 

113.1 

15,000 

96,750 

8,953,000 

9,124 

9.124 

129.7 

16,000 

103,200 

10,170,000 

10,890 

10.89 

147.7 

17,000 

109,650 

11,500,000 

11,720 

11.72 

166.6 

18,000 

116,100 

12,890,000 

18,140 

18.14 

186.8 

18,000 

122,550 

14,630,000 

14,680 

14.68 

208.1 

20,000 

129,000 

15,920,000 

16,280 

16.28 

280* 

14S 

2.IL— PERMEABILITY    OF  WROUGHT   AND    CAST   IRON. 

SQUARE  CENTIMETER  MEASUREMENT. 


Annealed  Wrought  Iron. 

Gray  Cast  Iron. 

B 

u 

H 

B 

J* 

H 

5,000 

8,000 

1.66 

4,000 

800 

5 

9,000 

2,250 

4 

5,000 

500 

10 

10,000 

2,000 

5 

6,000 

279 

21  5 

11,000 
12,000 
13,000 
14,000 
15,000 
16,000 

1,692 
1,412 

1,083 
823 
526 
820 

6.5 
8.5 
12 
17 
28.5 
50 

7,000 
8,000 
9,000 
10,000 
11,000 

133 
100 
71 
63 
87 

42 

80 
127 
183 
292 

17,000 

161 

105 

18,000 

90 

200 

19,000 

54 

850 

20,000 

80 

666 

SQUARE  INCH  MEASUREMENT. 


Annealed  "Wrought  Iron. 

Gray  Cast  Iron. 

B. 

/c, 

H. 

B. 

/*- 

H. 

80,000 
40,000 
50,000 
CO.OCO 
70,000 
80.000 
90,000 
100,000 
110,000 
120,000 
180,000 
140,000 

4,660 
8,877 
8,081 
2,159 
1,921 
1,409 
907 
408 
166 
76 
85 
27 

6.5 
10.3 
16.5 
27.3 
86.4 
56.8 
99.2 
245 
664 
1,581 
8,714 
5,185 

25,000 
80,000 
40,000 
60,000 
60,000 
70,000 

763 
756 
258 
114 
74 
40 

32.T 
89.7 
155 
439 
807 
1,480 

150 

PERMEABILITY  OP  SOFT  CHARCOAL  WROUGHT  IRON 

(8II1LFOBD   BITJWELL.) 
BQUABE   CENTIMETER  MEASUBE. 


B 

t* 

H 

7,890 
11,560 
15,460 
17,880 
18,470 
19,880 
19,820 

1899.1 
1121.4 
886.4 
150.7 
88.8 
45.8 
88.9 

8.9 
10.8 
40 
115 
208 
427 
585 

BQUAKE  INCH   MIA8UBEM1NT. 

B. 

//, 

H. 

47,414 
74,104 
99,191 
111,189 
118,504 
124,021 
127,165 

1897 
1122 
888 
150 
88.8 
45.8 
88.9 

25.0 
60.1 
256 
788 
1885 
2740 
8758 

B  —  Magnetic  Flux.         )  _ 

VBoth  in  lines  of  force. 
H  —  Magnetizing  Force.    J 

H  —  —  the  Permeability  or  multiplying  power  of  the  cor«. 


151 

XIII.-MAGNETIC   BELUCTANCE  OF  AIE  BETWEEN  TWO  PAEALLEL 
CYLINDEES  OF  IBON. 


b 

p 

Batio    of    least 

CBNTOTETKB  UNITS. 

INCH  UNITS. 

distance  apart 

to  circumference. 

0.1 

.1954 

5.1055 

0.0771 

12.968 

0.2 

.2707 

8.6917 

0.1066 

9.877 

0.8 

.8251 

8.0768 

0.1280 

7.815 

0.4 

.8688 

2.7158 

0.1450 

6.897 

0.5 

.4046 

2.4716 

0.1593 

6.278 

0.6 

.4861 

2.2983 

0.1717 

5.825 

0.8 

.4887 

2.0465 

0.1924 

5.198 

1.0 

.5816 

1.8807 

0.2093 

4.777 

1.2 

.5684 

1  .7996 

0.2238 

4.571 

1.4 

.6007 

1.6645 

0.2365 

4.228 

1.6 

.6289 

1.5902 

0.2476 

4.089 

1.8 

.6541 

1.5287 

0.2575 

8.883 

2.0 

.6774 

1.4764 

0.2667 

8.750 

4.0 

.8857 

1.1968 

0.3290 

8.040 

6.0 

.9319 

1.0782 

0.3669 

2.726 

8.0 

1.0047 

.9958 

0.3955 

2.528 

10.0 

1.0544 

.9484 

0.4151 

2.409 

In  this  table  in  columns  2  and  3  the  Unit  length  of  a  cylinder  is  taken  as  1  centi- 
meter ;  in  columns  4  and  5  as  1  inch,  p  —  circumference  of  cylinder  b  =  shortest 
distance  apart. 

XIV.-TABLE  OF  6TH  BOOTS. 


Num- 
ber 

Sixth 
Boot 

Number 

Sixth 
Boot 

Num- 
ber 

Sixth 
Boot 

Number 

Sixth 
Boot 

1 

.69355 

i 

.95320 

1* 

1.0177 

ii 

1.0978 

I 

.70717 

1 

.96350 

H 

1.0192 

If 

1.1019 

f 

.72306 

1 

.97006 

1* 

1.0226 

if 

1.1063 

I 

.74185 

f 

.97463 

u 

1.0260 

i? 

1.1087 

i 

.76473 

i 

.97798 

i| 

1.0308 

1$ 

1.1107 

i 

.79370 

1 

.98055 

i* 

1.0879 

if 

1.1119 

i 

.88263 

ft 

.98258 

n 

1.0491 

ift 

1.1129 

i 

.89090 

ii 

1.0^99 

2 

1.1237 

1 

.93462 

if 

1.0888 

152 

XV.-STANDARD   AND   BIRMINGHAM   WIRE  GAUGES. 


STANDARD. 

BIRMINGHAM. 

Number  of 
Gauge. 

Diameter 
in  Mils. 

Square  of 
Diameter  or 
Circ'l'r  Mils. 

Number  of 
Gauge. 

Diameter 
in  Mils. 

Square  of 
Diameter  or 
Circ'l'r  Mils. 

0000000 

500 

250000 

0000 

454 

206116 

000000 

464 

215296 

000 

425 

180625 

00000 

432 

186824 

00 

880 

144400 

0000 

400 

160000 

0 

840 

115600 

000 

372 

138384 

1 

800 

90000 

00 

348 

121104 

2 

284 

80656 

0 

324 

104976 

3 

259 

67081 

1 

800 

90000 

4 

238 

56644 

2 

276 

76176 

5 

220 

48400 

8 

252 

63504 

6 

203 

41209 

4 

282 

53824 

7 

180 

82400 

5 

212 

44944 

8 

165 

27225 

6 

192 

36864 

9 

148 

21904 

T 

176 

80976 

10 

184 

17956 

8 

160 

25600 

11 

120 

14400 

9 

144 

20736 

12 

109 

11881 

10 

128 

16384 

13 

095 

9025 

11 

116 

13456 

14 

083 

6889 

12 

104 

10816 

15 

072 

5184 

18 

092 

8464 

16 

065 

4225 

14 

080 

6400 

17 

058 

3864 

15 

072 

5184 

18 

049 

2401 

16 

064 

4096 

19 

042 

1764 

IT 

056 

8136 

20 

035 

1225 

18 

048 

2304 

21 

032 

1024 

19 

040 

1600 

22 

028 

784 

20 

086 

1296 

23 

025 

625 

21 

032 

1024 

24 

022 

484 

22 

028 

784 

25 

020 

400 

28 

024 

576 

26 

018 

324 

24 

022 

484 

25 

020 

400 

26 

018 

824 

153 


3    S    8 


3    8 


i  1 1 


s  s 


ill 


CO       ^       t-. 


154 


XVII.-WATTS  AND  HORSE  POWER  TABLES  FOR  VARIOUS 
PRESSURES  AND  CURRENTS. 

These  tables  will  be  found  very  convenient  for  quickly  finding  the 
watts  and  electrical  horse  power  on  lighting  and  power  circuits. 

To  find  the  watts  or  h.  p.  for  any  current  up  to  1,000  amperes  at  a 
standard  voltage  add  the  watts  or  h.  p.  corresponding  to  the  units, 
tens  and  hundreds  digits  of  the  current. 

Example :  Find  the  electrical  h.  p.  of  436  amperes  at  10$  volts. 

Solution:  The  h.  p.  for  400  amp.  is  56.3,  for  30  amp.  it  is  4.22  and  for 
6  amp.,  .845.  Adding  these  quantities  gives  61.365  h.  p.,  which  we  will 
call  61.4  h.  p.,  as  the  tabular  values  are  computed  to  three  figures 
only,  which  are  sufficient  for  engineering  purposes. 

To  find  values  for  voltages  higher  or  lower  than  in  the  tables, 
select  a  voltage  1-10  or  ten  times  that  required,  and  multiply  the  re- 
sult by  10  or  1-10.  Thus :  to  find  h.  p.  at  7  amp.,  fi5  volts,  take  7  amp. 
at  550  volts*5.16  h.  p.;  multiply  by  1-10,  which  gives  .516  h.  p.  To 
find  h.  p.  at  9  amp.,  i,200  volts,  take  9  amp.  at  120  volts=1.45 ;  multi- 
ply by  10=14  5  h.  p. 

To  read  in  kilowatts  place  a  decimal  point  before  the  watts  when 
less  than  1,000  in  value,  or  substitute  it  for  the  comma  in  the  larger 
values, 

HORSE  POWER  AT  VARIOUS  PRESSURES  AND  CURRENTS. 


100  volts. 

105  volts. 

110  volts. 

Amperes. 

Watts. 
100 

Mi 

Watts. 
105 

h.p. 

.141 

Watts. 
110 

h.p. 

.147 

2 

200 

.268 

210 

.282 

220 

.295 

3 

300 

.402 

315 

.422 

330 

.442 

4 

400 

.636 

420 

.563 

440 

.590 

5 

500 

.670 

525 

.704 

550 

.737 

6 

600 

.804 

630 

.845 

660 

.885 

7 

700 

.938 

735 

.985 

770 

1.03 

8 

800 

1.07 

840 

1.13 

880 

1.18 

9 

900 

1.21 

945 

1.27 

990 

1.33 

10 

1,000 

1.34 

1,050 

1.41 

1,100 

1.47 

20 

2,000 

2.68 

2,100 

2.82 

2,2uO 

2.95 

30 

3,000 

4.02 

3,150 

4.22 

3,300 

4.42 

40 

4,000 

5.36 

4,200 

563 

4,400 

5.90 

50 

5000 

6.70 

5,250 

7.04 

5,500 

7.37 

60 

6,000 

8.04 

6,300 

8.45 

6,600 

8.85 

70 

7,000 

9.38 

7,350 

9.85 

7,700 

10.3 

80 

8,000 

10.7 

8,400 

11.3 

8,800 

11.8 

90 

9,000 

12.1 

9,450 

12.7 

9,9(10 

13.3 

100 

10,(K)0 

13.4 

10,500 

14.1 

11,000 

14.7 

.   200 

20.000 

28.8 

21,000 

28.2 

22,000 

29.5 

300 

30,000 

40.2 

31,500 

42.2 

33000 

44.2 

400 

40,000 

53.6 

42,000 

56.3 

44,000 

59.0 

500 

50,000 

67.0 

52,500 

70.4 

55,000 

73.7 

600 

60,000 

80.4 

63,000 

84.5 

66,000 

88.5 

700 

70,000 

93.8 

73,500 

98.5 

77,000 

103 

800 

80,000 

107 

84,000 

113 

88,000 

118 

900 

90,000 

121 

94,500 

127 

99,000 

133 

1,000 

100,000 

134 

105,000 

141 

110,000 

147 

155 


HORSE  POWER  AT  VARIOUS  PRESSURES  AND  CUBBENTS. 
(Continued.) 


115  volts. 

120  volts. 

125  volts. 

Amp. 

Watts. 
115 

Mi 

Watts. 
120 

h.  p. 
.161 

Watts. 
125 

h.p. 
.168 

2 

230 

.308 

240 

.322 

25') 

.335 

3 

345 

.462 

36«> 

.483 

375 

.505 

4 

460 

.617 

480 

.644 

»0 

.670 

5 

575 

.770 

6  0 

.805 

625 

.840 

6 

690 

.925 

730 

.fc66 

750 

1.01 

7 

805 

1.08 

840 

1.13 

875 

1.18 

8 

920 

1.23 

96» 

1.29 

1,000 

1.34 

9 

1,035 

1.39 

1,<>80 

145 

1,125 

1.51 

10 

1,150 

1.54 

1.200 

161 

1,250 

168 

20 

2.300 

3.08 

2,400 

3.22 

2.500 

3.35 

30 

3,450 

4.62 

3,600 

483 

3,750 

5.C5 

40 

4,600 

6.17 

4,#iO 

6.44 

S.rOO 

6.70 

50 

5,750 

7.70 

6,(KIU 

8.04 

6.250 

8.40 

60 

6,900 

9.25 

7,200 

966 

7.5  0 

10.1 

70 

8.050 

10.8 

8,400 

11.3 

8,750 

11.8 

80 

9.200 

12.3 

9,600 

12.9 

10  POO 

13.4 

90 

10350 

13.9 

1K800 

145 

11,250 

15.1 

100 

11,500 

154 

12,(00 

16.1 

12,500 

16.8 

200 

23,  (00 

30.8 

24.'  00 

32.2 

25,000 

335 

300 

34,500 

462 

36,000 

483 

37,500 

505 

400 

46/00 

617 

48,000 

64.4 

50.(00 

67.0 

500 

57.500 

77.0 

60(K<0 

80.4 

62,500 

840 

600 

69.(00 

92.5 

72,i  M) 

96.6 

75,000 

1)1 

700 

80500 

lf'8 

84,»0 

113 

87,5<0 

118 

800 

92,000 

123 

96,™  > 

129 

100.000 

134 

900 

103,  00 

139 

118,000 

145 

112.50«) 

151 

1,000 

115,000 

154 

120,000 

161 

125,ttO 

168 

156 


HORSE  POWER  AT  VARIOUS  PRESSURES  AND  CURRENTS. 
(Continued.) 


200  volts. 

210  volts. 

220  volts. 

Amp. 

Watts. 
200 

h.p. 

.268 

Watts. 
210 

h.p. 

.282 

Watts. 
220 

h.p. 

.295 

2 

400 

.536 

420 

.563 

440 

.690 

3 

600 

.8U4 

630 

.845 

660 

.885 

4 

800 

1.07 

840 

1.13 

88'  1 

1.18 

5 

1,000 

1.34 

1,050 

1.41 

1,100 

1.47 

6 

1,200 

1.61 

1,260 

1.69 

1,320 

1.77 

7 

1,400 

1.88 

1,470 

1.97 

1,540 

2.06 

8 

1,600 

2.14 

1.680 

2.25 

1.760 

2.36 

9 

1,800 

241 

1,890 

253 

1,980 

2.65 

10 

2,000 

2.68 

2,100 

2.82 

2,200 

2.95 

20 

4,000 

5.36 

4,200 

5.63 

4,400 

5.90 

30 

6.UOO 

8.04 

6,300 

845 

6,600 

8.85 

40 

8,000 

10.7 

8,400 

11.3 

8,800 

11.8 

50 

li»,000 

13.4 

10,500 

14.1 

11,000 

14.7 

60 

12,CM) 

16.1 

12,600 

16.9 

13,200 

177 

70 

14,0(0 

18.8 

14,700 

19.7 

15,400 

20.6 

80 

16,000 

21.4 

16.800 

22.5 

17,600 

23.6 

90 

18,000 

241 

18,900 

26.3 

19,800 

26.5 

100 

20,000 

26.8 

21,000 

28.2 

22,000 

295 

200 

40,000 

53.6 

42,000 

56.3 

44,000 

59.0 

300 

60,000 

80.4 

63,000 

84.5 

66,000 

88.5 

400 

80,000 

107 

84,000 

113 

88,000 

118 

500 

lOO.OCO 

134 

105,000 

141 

110,0(  0 

147 

600 

120,000 

161 

126,000 

169 

132,000 

177 

700 

140,00) 

188 

147,000 

197 

154,000 

206 

800 

160,000 

214 

168,000 

225 

176,000 

236 

900 

180,000 

241 

189,000 

253 

198,000 

881 

1,OUO 

200,000 

288 

210,000 

282 

220,000 

295 

157 


HORSE  POWER  AT  VARIOUS  PRESSURES  AND  CURRENTS. 
(Continued.) 


230  volte. 

240  volts. 

250  volts. 

Amp. 

Watts. 
230 

":& 

Watts. 
240 

h-p. 
.322 

Watts. 
250 

h.p. 
.&5 

2 

460 

.617 

480 

.644 

500 

.670 

3 

690 

.925 

720 

.966 

750 

1.01 

4 

920 

1.23 

960 

1.29 

1,000 

1.34 

5 

1,150 

1.54 

1,200 

1.61 

1,250 

1.68 

6 

1,380 

1.85 

1,440 

1.93 

1,500 

2.01 

7 

1,610 

2.16 

1,680 

2.25 

1,760 

235 

8 

1,840 

2.47 

1,920 

2.58 

2,000 

2.68 

9 

2,070 

2.77 

2,160 

2.90 

2,2AO 

3.02 

10 

2,300 

3.08 

2,400 

3.22 

2,500 

335 

20 

4,600 

6.17 

4,800 

6.44 

5.000 

6.70 

30 

6,900 

9.*5 

7,200 

9.66 

7,500 

10.1 

40 

9,200 

12.3 

9,600 

12.9 

10.000 

13.4 

50 

11,500 

15.4 

12,000 

16.1 

12,500 

168 

60 

13,800 

185 

1440) 

19.3 

15,000 

20.1 

70 

16,100 

21.6 

16,800 

22.5 

17,500 

23.6 

80 

18,400 

24.7 

19,200 

258 

20,000 

26.8 

90 

20,700 

27.7 

21,600 

29.0 

22,500 

30.2 

100 

23.000 

30.8 

24,000 

32.2 

20.000 

33.6 

no 

46,000 

61.7 

48000 

64.4 

50,000 

67.0 

300 

69000 

92.5 

72,000 

96.6 

75.000 

101 

400 

98.000 

123 

96.1  111 

129 

100,000 

134 

500 

115,000 

154 

120,000 

161 

1  5000 

168 

600 

138,000 

185 

144,000 

193 

150.000 

201 

700 

161.COO 

216 

168.000 

225 

175,01  0 

236 

800 

184.000 

247 

192,000 

258 

2<UOOO 

268 

900 

aiTloro 

277 

216,000 

290 

225.000 

302 

1,000 

230,000 

308 

240,000 

322 

250,000 

335 

L53 


HORSE  POWER  AT  VARIOUS  PRESSURES  AND  CURRENTS. 
(Continued.) 


SCO  volts. 

550  volts. 

600  volts. 

Watts 

Watts 

Watts 

Amp. 

and  k.  w. 

500 

h'.& 

and  k.  w. 

5.'0 

h:ft 

and  k.  w. 

eoo 

.804 

2 

1,000 

1.34 

1,1(0 

1.47 

1,200 

1.61 

3 

1,500 

201 

1,650 

2.21 

1,800 

2.41 

4 

2,000 

2.68 

2,200 

2.95 

2,400 

322 

5 

2,500 

3.35 

8.7fO 

3.69 

3.000 

402 

6 

3,000 

4.02 

3,3(0 

442 

8,600 

4.82 

7 

3,500 

4.69 

3,850 

5.16 

4200 

5.63 

8 

4,000 

5.36 

4,400 

590 

4800 

643 

9 

4,500 

6.03 

4.9oO 

6.64 

5,400 

724 

10 

5k.w. 

6.7 

5.5  k.w. 

7.4 

6  k.w. 

8.04 

20 

10 

13.4 

11.0 

14.7 

12 

16.1 

30 

15 

201 

16.5 

221 

18 

24.1 

40 

20 

268 

22.0 

29.5 

24 

32.2 

50 

25 

33,5 

27.5 

36.9 

30 

40.2 

60 

30 

40.2 

33.0 

44.2 

36 

48.2 

70 

35 

469 

38.5 

51.6 

42 

66.3 

80 

40 

636 

44.0 

59.0 

48 

64.3 

90 

45 

60.3 

49.5 

66.4 

54 

72.4 

100 

50 

67 

55 

74.0 

60 

80.4 

200 

100 

134 

110 

147 

120 

161 

300 

150 

201 

165 

221 

180 

241 

4l,0 

2(10 

268 

220 

295 

240 

322 

500 

260 

33S 

275 

369 

300 

402 

600 

300 

402 

330 

442 

360 

483 

700 

'350 

469 

J-85 

616 

420 

563 

800 

400 

536 

440 

590 

480 

643 

900 

450 

603 

495 

664 

540 

724 

1,OUO 

500 

670 

550 

740 

600 

804 

INDEX. 


ALTERNATING  current  sys- 
tem   47-49 

Alternating   currents    115 

Amperage  of  armature...     99 

Ampere    11-12 

Ampere-second     57 

Ampere-turns    87 

Ampere-turns,     rules     for 

calculating 87 

Armature  amperage  of,  on 

short  circuit   99 

Armature   calculations.  .99-103 
Armature,  capacity  of.  ...      99 
Armature,      general      fea- 
tures of 96 

Armature,  resistance  of . .     98 
Armatures,    rules   for   cal- 
culating   98-99 

Armature,   voltage   of. ...      99 
Armature  winding  iron  or 
copper 96 

BRIDGE,  Wheatstone,  prin- 
ciple of 25 

Batteries,  illustrations  of 
arrangements  66 

Batteries  or  generators  in 
opposition  15 

Batteries,  storage,  resist- 
ance of 65 

Battery  arrangement  and 
size  for  given  efficiency.  73 

Battery,  arrangement  of 
cells  in 67 

Battery  calculations,  dis- 
crepancies in 72 

Battery,  chemicals  con- 
sumed in 78 

Battery    constants 65 

Battery,  current  of 68 

Battery,  effective  rate  of 
work  of  77-78 


Battery,  efficiency,  to  cal- 
culate    72 

Battery,  electromotive 
force  of  67 

Battery,  how  rated 65 

Battery,  resistance  of....     67 

Battery,  to  calculate  cur- 
rent and  arrangement. 69-72 

Battery,  to  calculate  its 
voltage 76-77 

Battery,  work  of 77-78 

CALORIE,  gram  and  kilo- 
gram, defined 54 

Calorie,  relation  to  watts 

and  ergs 54-55 

Capacity       of       armature 

winding    97 

Cars,  power  to  move  ....   Ill 

Cell   constants 65 

Chemicals  consumed  in  a 

battery    78 

Circuits,   divided,  branched 

or  shunt   19-25 

Circuits,   portions   of....  17-19 
Circuits,  single  conductor, 

closed 14-15 

Circular  mils  calcula- 
tions   44-45 

Circular  mils 43-45 

Condensers 121 

Conductance    36-37 

Conductors  of  same  mate- 
rial    26 

Conductors,    resistance    of 

different   26 

Contact,  area  of  in  elec- 
tro-magnets and  arma- 
tures    86 

Convection  and   radiation, 

loss  of  heat  by 58 

Conversion,   ratio  of 4J 


160 


INDEX. 


Converters,  rule  for  wind- 
ing    49 

Cores  of  field-magnet,  to 
calculate  104 

Counter-electromotive  force 
of  plating-bath 80 

Current,  battery  required 
for  a  given 69-72 

Current,  distribution  in 
parallel  circuits 20-21 

Current,  its  heating  effect 
and  energy 50 

Currents  passed  by 
branches  of  a  circuit. .  19 

Current,  work  of 60 

Current  yielded  by  a  bat- 
tery    65 

DEFINITIONS,,  general 9 

Demonstrations   of  rules.   127 

Derived  units   10 

Dimensions,  units  of 10 

Drop      of     potential      in 

leads    38-41 

Drum  type    closed    circuit 

armatures 98-99 

Duty  and  commercial  ef- 
ficiency   63-64 

Duty  of  generators 63 

EFFECTIVE  B.  M.  F 16 

Efficiency,  its  relation  to 
resistances  64 

Efficiency  of  generators, 
commercial  64 

Efficiency  of  generators, 
electrical  63 

Efficiency,  to  calculate 
battery  for  a  given .  .  73-74 

Electrical   railways    109 

Electro-magnets  and  dy- 
namos    82 

Electro-magnets,  general 
rules  for  85 

Electro-magnets,  to  mag- 
netize    86-87 

Electro-magnets,  traction 
of  85-86 

Electromotive  force  of 
battery  67 

Electro  -  plating    calcula- 
tions          79 

Energy    defined     9 

Energy  in  circuit,  rules 
for  determining 50-53 

Energy  of  current 50-53 


E.  M.  F.,  its  meaning. . .  10 
E.  M.  F.,  primary  and  sec- 
ondary      47 

Erg   54 

FORCE  defined    9 

Force,    magnetic    84 

Fundamental    units 10 

Field-magnet  cores,  to  cal- 
culate    104 

Field-magnet   for    dynamo 

or  motor  104-107 

Field,  unit  intensity  of 
magnetic 82-83 

GENERATORS,  efficiency  of, 

63-64 

Generators  or  batteries  in 
opposition  15 

HEAT,  absolute  quantity  in 

circuit    54 

Heating  effect  of  current, 

50-53 

Heat,  specific 57 

Horse-power,    electrical . .     61 
Horse  -  power,       reduction 

from    kilogram-meters.  .      60 
Horse-power,    to   calculate 
for   lamps 62 

JOULE,  his  law  of  heating, 

effect  of  currents 50 

Joule  or  gram-calorie. ...     54 

KAPP,  Gisbert,  his  line  of 

force 107-108 

Kilogram-meters    60 

LEADS,  tapering  in  size.  .  41-42 
Leakage  between  cylindri- 
cal  magnet   leg,  to  cal- 
culate        91 

Leakage  between  flat  mag- 
net surfaces,  to  calcu- 
late    91 

Leakage  of  lines  in  a  mag- 
netic circuit  95 

Leakage  of  lines  of  force.     89 
Line  of  force,  Kapp's..  107-108 

Lines  of  force    82 

Lines  of  force  cut  per  sec- 
ond for  one  volt 96 

Lines  of  force,  leakage  of.     89 
Lines  of  force,  to  diminish 
leakage  of 94-95 


INDEX. 


161 


MAGNETIC  circuit 84 

Magnetic    circuit    calcula- 
tions     88-89 

Magnetic  circuit,   the   law 

of    85 

Magnetic      circuits,      four 

parts  of 88 

Magnetic  circuits,  general 

calculations  for    92 

Magnetic    field    82-83 

Magnetic  flux   82-83 

Magnetic  force    84 

Magnetic    potential,    aver- 
age difference   of 90 

Magnetism,  no  insulator  of     84 

Magnetizing  force 87 

Magnet     legs,     long     and 

short 94 

Mass    defined    9 

Metals,    deposition    of,   by 

battery 79 

Mho,   unit  of  conductance     36 
Mil,    circular,   calculations 

based   on    44-45 

Mils,   circular    43 

Mils,    circular,   applied   to 

alternating  current    ...      48 
Multiple    arc    connections, 
to    calculate    38-42 

NEUTRAL    wire     in     three 

wire   system    46 

Notation  in  powers  of  ten  136 

OHM   11-12 

Ohm's    law    13-25 

Ohm's    law,    its    universal 

application     14 

Ohm's     law,     six     expres- 
sions of   13-14 

PARALLEL    connections    of 

equal     resistance 23 

Parallel    leads,    resistance 

of    22 

Permeance   83-84 

Permeability    85 

Permeability,     average 

range     of     104 

Potential,  diagram  for  cal- 
culating   fall    of 41 

Potential  difference   ....  38-42 
Potential  difference,  drop, 

or    fall    of 17 

Potential,  drop  of  in  leads, 

38-42 


Power  to  move  cars Ill 

Powers  of  ten,  notation  in  136 

Primary  B.  M.  F 47 

Proving  armature  calcula- 
tions         103 

RADIATION  and  convection, 

loss   of   heat   by 58 

Railways,  electric 1U9 

Rate  of  heat-energy,  units 

of    63 

Ratio  of  conversion 47 

Reluctance    83-84 

Reluctance,  calculation  of.     87 

Resistance    26-35 

Resistance   and    efficiency, 

how    related    64 

Resistance  defined    9 

Resistance  of  battery....      67 
Resistance  of  circuit,  note 

relative   thereto    14 

Resistance       of       parallel 

leads    22-23 

Resistance       referred       to 
weight  of  conductor. .  .35-36 

Resistance  specific 29 

Resistances,    two    in    bat- 
tery  circuits 65 

Resistance,   universal   rule 

for    31-32 

Rules,  demonstrations  of.    127 

SAFETY-CATCHES  for  fuses, 

how  calculated   59 

Secondary   E.   M.   F 47 

Self-Induction    117 

Series    winding   for   dyna- 
mos       106 

Shunt  circuits      19 

Shunt  circuits,   resistance 

of    22 

Shunt   winding   for    dyna- 
mos       107 

Sizes  of  feeders 109 

Space    defined 9 

Specific  heat 57 

Specific   resistance 30-31 

System,     alternating    cur- 
rent     47-49 

Systems,    special 46-49 

System,  three  wire 46-47 

TABLES — 

American       wire       gauge 
table 146 


162 


INDEX. 


TABLES — Continued. 

Chemical  and  electro- 
chemical equivalents  .  .  148 

Chemical  and  thermo- 
chemical  equivalents  .  .  147 

Current  capacity  of  bare 
or  insulated  overhead 
wires 153 

Equivalents  of  units  o± 
area  140 

Equivalents  of  units  of 
energy  and  work .  . .  142-143 

Equivalents  of  units  of 
length  139 

Equivalents  of  units  of 
volume  140 

Equivalents  of  units  of 
weight  141 

Magnetic  reluctance  of  air 
between  two  parallel 
cylinders  of  iron  151 

Magnetization  and  mag- 
netic traction 148 

Permeability  of  soft  char- 
coal wrought  iron  ....  150 

Permeability  of  wrought 
and  cast  iron  149 

Relative  resistance  and 
conductance  of  pure 
copper  at  different  tem- 
peratures    145 

Specific  resistance  of  solu- 
tions and  liquids 144 

Standard  and  Birmingham 
wire  gauges 152 

Table  of  specific  resist- 
ances in  microhms  and 


of  coefficients  of  specific 

resistances  of  metals.  .  144 

Table  of  the  sixth  roots.  .  151 

Three  wire  system 48 

Three  wire  system,  saving 

in  size  of  wires 46 

Three  wire  system,  the 

neutral  wire 46 

Time  defined 9 

UNITS,  concrete  statement 

of 12 

Units,   fundamental 10 

Units,  original  and  de- 
rived    10 

VOLT     10-12 

Voltage  of  battery,  to  cal- 
culate   76-77 

Volt  amperes .' 56-60 

WATT   56-60 

Watts  and  horse  power 
tables  for  various  pres- 
sures and  currents.  .  .  .  154 

Weight  defined 9 

Weight    of    conductor,    re- 
sistance  referred  to... 35-36 
Wheatstone    bridge,     prin- 
ciple   of 25 

Winding,        series        and 

shunt    106-107 

Wire,  rule  for  heating  of, 

by  a  current 59 

Work  defined 9 

Work  of  current 60 


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BAUER.   Marine  Engines  and  Boilers:   Their  De- 
sign and  Construction. 

A  large  practical  work  of  722  pages,  550  illustra- 
tions, and  17  folding  plates,  for  the  use  of  students, 
engineers  and  naval  constructors.  $9.00  net. 

BAXTER,  JR.    Switchboards. 

The  only  book  dealing  with  this  important  part  of 
electrical  engineering.  Takes  up  all  sizes  and  kinds 
from  the  single  dynamo  in  the  engine  room  to  the 
largest  power  plant  work.  Includes  divert  and  alter- 
nating currents;  oil  switches  for  high  tension;  arc  and 
incandescent  lighting;  railway  work,  and  all  the  rest, 
except  telephone  work.  $1.50. 

BAXTER,  JR.    Commutator  Construction. 

The  business  end  of  a  dynamo  or  motor  is  the  com- 
mutator, and  this  is  what  is  apt  to  give  trouble.  This 
shows  how  they  are  made,  why  they  get  out  of  whack 
and  what  to  do  to  put  'em  right  again.  Price,  25 
cents. 

BENJAMIN.    Modern  Mechanism. 

A  practical  treatise  on  machines,  motors,  and  the 
transmission  of  power,  being  a  complete  work  and  a 
supplementary  volume  to  Appleton  s  Cyclopedia  of 
Applied  Mechanics.  Bound  in  half  Morocco.  Price, 
$5.00. 

BLACKALL.    Air-Brake  Catechism. 

This  book  is  a  complete  study  of  the  air-brake 
equipment,  including  the  E.  T.  Equipment,  as  well  as 
all  the  latest  devices.  All  parts  of  the  air  brake,  their 
troubles  and  peculiarities,  and  a  practical  way  to  find 
and  remedy  them,  are  explained.  This  book  contains 
over  1,500  questions  with  their  answers.  312  pages. 
$2.00. 

BLACKALL.   New  York  Air-Brake  Catechism. 

This  is  a  complete  treatise  on  the  New  York  Air- 
Brake  and  Air-Signalling  Apparatus,  giving  a  detailed 
description  ot  all  the  parts,  their  operation,  troubles, 
and  the  methods  of  locating  and  remedying  the  same. 
200  pages,  fully  illustrated.  $1.00. 

BOOTH  AND  KERSHAW.    Smoke  Prevention  and 

Fuel  Economy. 

As  the  title  indicates,  this  book  of  197  pages  and 
75  illustrations  deals  with  the  problem  of  complete 
combustion,  which  it  treats  from  the  chemical  and 
mechanical  standpoints,  besides  pointing  out  the  eco- 
nomical and  humanitarian  aspects  of  the  question. 
$2.50. 


GOOD,  PRACTICAL  BOOKS 


BOOTH.    Steam  Pipes:  Their  Design  and  Construc- 
tion. 

A  treatise  on  the  principles  of  steam  conveyance  and 
means  and  materials  employed  in  practice,  to  secure 
economy,  efficiency,  and  safety.  A  book  which  should 
be  in  the  possession  of  every  engineer  and  contractor. 
$2.00. 

BUCHETTI.  Engine  Tests  and  Boiler  Efficiencies. 
This  work  fully  describes  and  illustrates  the  meth- 
od of  testing  the  power  of  steam  engines,  turbine  and 
explosive  motors.  The  properties  of  steam  and  the 
evaporative  power  of  fuels.  Combustion  of  fuel  and 
chimney  draft;  with  formulas  explained  or  practically 
computed.  255  pages.  179  illustrations.  $3.00. 

BYRON.    Physics  and  Chemistry  of  Mining. 

For  the  use  of  all  preparing  for  examinations  in 
Mining  or  qualifying  for  Colliery  Managers'  Certifi- 
cates. $2.00. 

Car  Charts. 

Shows  and  names  all  the  parts  of  three  types  of 
cars.  Passenger — Box-gondola.  Printed  on  heavy 
plate  paper  and  mailed  in  a  tube.  25  cents  each.  Set 
of  3  for  50  cents. 

COCXIN.    Practical  Coal  Mining. 

An  important  work,  containing  428  pages  and  213 
illustrations,  complete  with  practical  details,  which  will 
intuitively  impart  to  the  reader,  not  only  a  general 
knowledge  of  the  principles  of  coal  mining,  but  also 
considerable  insight  into  allied  subjects.  The  treatise 
is  positively  up  to  date  in  every  instance,  and  should 
be  in  the  hands  of  every  colliery  engineer,  geologist, 
mine  operator,  superintendent,  foreman,  and  all  others 
who  are  interested  in  or  connected  with  the  industry. 
$2.50. 

COLVIN.    American  Compound  Locomotives. 

The  latest  and  most  complete  book  on  compounds. 
Shows  £ll  types,  including  the  balanced  compound 
which  is  now  being  used.  Makes  everything  clear  by 
many  illustrations,  and  shows  valve  setting,  break- 
downs and  repairs.  $1.50.  v 

COLVIN  AND  CHENEY.    Engineer's  Arithmetic. 

A  companion  to  Machine  Shop  Arithmetic,  arranged 
for  the  stationary  engineer.  Shows  how  to  work  the 
problems  of  the  engine  room  and  shows  "why."  Has 
steam  tables  and  a  lot  of  other  useful  information 
that  makes  it  popular  with  practical  men.  50  cents. 


GOOD,  PRACTICAL  BOOKS 


COLVIN.   Link  Motions  and  Valva  Setting, 

A  handy  little  book  for  the  engineer  or  machinist 
that  clears  up  the  mysteries  of  valve  setting.  Shows 
the  different  valve  gears  in  use,  how  they  work  and 
why.  Piston  and  slide  valves  of  different  types  are 
illustrated  and  explained.  A  book  that  every  rail- 
road man  in  the  motive  power  department  ought  to 
have.  60  cents. 

COLVIN  AND  CHENEY.  Machine  Shop  Arithmetic. 

Most  popular  book  for  shop  men.  Shows  how  all 
shop  problems  are  worked  out  and  "why."  Includes 
change  gears  for  cutting  any  threads;  drills,  taps, 
shink  and  force  fits;  metric  system  of  measurements 
and  threads.  Used  by  all  classes  of  mechanics  and 
for  instruction  by  Y.  M.  C.  A.  and  other  schools. 
60  cents. 

COLVIN.    The  Railroad  Pocketbook. 

Different  from  any  book  you  ever  saw.  Gives  clear 
and  concise  information  on  just  the  points  you  are 
interested  in.  It's  really  a  pocket  dictionary,  fully 
illustrated,  and  so  arranged  that  you  can  find  just 
what  you  want  in  a  second  without  an  index.  Whether 
you  are  interested  in  Axles  or  Acetylene;  Compounds 
or  Counter  Balancing;  Rails  or  Reducing  Valves; 
Tires  or  Turntables,  you'll  find  them  in  this  little 
book.  It's  very  complete.  Flexible  cloth  cover.  200 
pages.  $1.00.  Interleaved  with  ruled  pages  for  notes. 
$1.50. 

COLVIN-STABEL.    Threads  and  Thread  Cutting:. 

This  clears  up  many  of  the  mysteries  of  thread- 
cutting,  such  as  double  and  triple  threads,  internal 
threads,  catching  threads,  use  of  hobs,  etc.  Contains 
a  lot  of  useful  hints  and  several  tables.  Price,  25 
cents. 

COLVIN.    Turning  and  Boring  Tapers. 

A  plainly  written  explanation  of  a  subject  that  puz- 
zles many  a  mechanic.  This  explains  the  different 
ways  of  designating  tapers,  gives  tables,  shows  how  to 
use  the  compound  rest  and  gives  the  tapers  mostly 
used.  Price,  25  cents. 
CRANE.  American  Stationary  Engineering. 

A  new  book  by  a  well-known  author.  Begins  at  the 
boiler  room  and  takes  in  the  whole  power  plant.  _  Con- 
tains the  result  of  years  of  practical  experience  in  all 
sorts  of  engine  rooms  and  gives  exact  information  that 
cannot  be  found  elsewhere.  It's  plain  enough  for  prac- 
tical men  and  yet  of  value  to  those  high  in  the  pro- 
fession. Has  a  complete  examination  for  a  license. 
19.00. 


GOOD,  PRACTICAL  BOOKS 


DALEY.   Train  Hules  and  Dispatching, 

Contains  the  standard  code  for  both  single  and 
double  track  and  explains  how  trains  are  handled  un- 
der all  conditions.  Gives  all  signals  in  colors,  is  illus- 
trated wherever  necessary,  and  the  most  complete  book 
in  print  on  this  important  subject.  Bound  in  fine  sea 
flexible  leather.  221  pages.  $1.50. 

EXGSTROM,    Bevel  Gear  Tables. 

No  one  who  has  to  do  with  bevel  gears  in  any  w&/ 
should  be  without  this  book.  The  designer  and  drafts- 
man will  find  it  a  great  convenience,  while  to  the 
machinist  who  turns  up  the  blanks  or  cuts  the  teeth, 
it  is  invaluable,  as  all  needed  dimensions  are  given 
and  no  fancy  figuring  need  be  done.  $1.00. 

FOWLER.    Boiler  Room  Chart. 

An  educational  chart  showing  in  isometric  perspec- 
tive the  mechanisms  belonging  in  a  modern  boiler- 
room.  Each  part  is  given  a  reference  number,  and 
these,  with  the  corresponding  name,  are  given  in  a 
glossary  printed  at  the  sides.  The  chart,  therefore, 
serves  as  a  dictionary  of  the  boiler-room,  the  names 
of  more  than  two  hundred  parts  being  given  on  the 
list.  25  cents. 

FOWLER.      Locomotive     Breakdowns     and     Their 

Remedies. 

This  work  treats  in  full  all  kinds  of  accidents  that 
are  likely  to  happen  to  locomotive  engines  while  on 
the  road.  The  various  parts  of  the  locomotives  are 
discussed,  and  every  accident  that  can  possibly  happen, 
with  the  remedy  to  be  applied,  is  given.  250  pages. 
$1.50. 

GODDARD.    Eminent  Engineers. 

An  intensely  interesting  account  of  the  achieve- 
ments of  thirty  of  the  world's  best  known  engineers. 
Free  from  tiresome  details  and  giving  just  the  facts 
you  want  to  know  in  an  entertaining  manner.  Por- 
traits are  given  and  the  book  is  an  inspiration  for 
both  old  and  young.  $1.50. 

GEIMSHAW.   Saw  Filing  and  Management  of  Saws. 

A  practical  handbook  on  filing,  gumming,  swaging, 
hammering,  and  the  brazing  of  band  saws,  the  speed, 
work,  and  power  to  run  circular  saws,  etc.,  etc.  $1.00. 
GRIMSHAW.  "Shop  Kinks." 

This  shows  special  methods  of  doing  work  of  vari- 
ous kinds,  and  reducing  cost  of  production.  Has 
hints  and  kinks  from  some  of  the  largest  shops  in 
this  country  and  Europe.  You  are  almost  sure  to  find 
some  that  apply  to  your  work,  and  in  such  a  way  to 
save  time  and  trouble.  400  pages.  Price,  $2.50.  A 


GOOD,  PRACTICAL  BOOKS 


GRIMSHAW.    Engine  Runner's  Catechism. 

Tells  how  to  erect,  adjust,  and  run  the  principal 
steam  engines  in  use  in  the  United  States.  336  pages, 
$2.00. 

GRIMSHAW.    Steam  Engine  Catechism. 

A  series  of  direct  practical  answers  to  direct  prac- 
tical questions,  mainly  intended  for  young  engineers 
and  for  examination  questions.  Nearly  1,000  questions 
with  their  answers.  413  pages.  $2.00. 

GRIMSHAW.    Locomotive  Catechism. 

This  is  a  veritable  encyclopedia  of  the  locomotive, 
is  entirely  free  from  mathematics,  and  thoroughly  up 
to  date.  It  contains  1,600  questions  with  their  an- 
swers. 450  pages,  over  200  illustrations.  $2.00. 

HARRISON.     Electric  Wiring,   Diagrams   and 
Switchboards. 

A  thorough  treatise  covering  the  subject  in  all  its 
branches.  Practical  every-day  problems  in  wiring  are 
presented  and  the  method  of  obtaining  intelligent 
results  clearly  shown.  270  pages.  105  illustrations. 
$1.50. 

Henley's    Twentieth    Century    Book    of    Receipts, 
Formulas  and  Processes. 

Edited  by  G.  D.  Hiscox.  The  most  valuable  Tech- 
no-Chemical  Receipt  Book  published.  Contains  over 
10,000  selected  scientific  chemical,  technological  and 
practical  receipts  and  processes,  including  hundreds  of 
so-called  trade  secrets  for  every  business.  900  pages. 
Price,  $3.00. 

HISCOX.    Gas,  Gasoline,  and  Oil  Engines. 

Every  user  of  a  gas  engine  needs  this  book.  Sim- 
ple, instructive,  and  right  up  to  date.  The  only  com- 
plete work  on  the  subject.  Tells  all  about  the  run- 
ning and  management  of  gas  engines.  Includes  chap- 
ters on  horseless  vehicles,  electric  lighting,  marine 
propulsion,  etc.  450  pages.  Illustrated  with  351  en- 
gravings. $2.50. 

Henley's  Encyclopedia  of  Practical  Engineering  and 
Allied  Trades. 


Edited  by  JOSEPH  G.  HORNER.  Complete  in  five 
alumes.  Each  volume  contains  500  pages  and  500 
illustrations.  Bound  in  half  morocco.  Price,  $6.00 


per  volume,  or   $25.00   for  the  complete  set  of  five 
volumes. 


GOOD,  PRACTICAL  BOOKS 


HISCOX.  Compressed  Air  in  All  Its  Applications. 
This  is  the  most  complete  book  on  the  subject  of 
Air  that  has  ever  been  issued,  and  its  thirty-five  chap- 
ters include  about  every  phase  of  the  subject  one  can 
think  of.  It  may  be  called  an  encyclopedia  of  com- 
pressed air.  It  is  written  by  an  expert,  who,  in  its 
825  pages,  has  dealt  with  the  subject  in  a  comprehen- 
sive manner,  no  phase  of  it  being  omitted.  545  illus- 
trations, 820  pages.  Price,  $5.00. 

HISCOX.    Mechanical  Movements,   Powers,   and 

Devices. 

Almost  every  mechanic  is  interested  in  mechanical 
movements,  and  the  designer  finds  new  problems  every 
day.  A  book  of  this  kind  used  as  a  reference  is  in- 
valuable in  solving  some  of  the  many  problems  in 
mechanics,  especially  in  designing  new  machinery. 
400  pages.  Price,  $3.00. 

HISCOX.    Mechanical  Appliances,  Mechanical  Move- 

ments  and  Novelties  of  Construction. 
This  is  a  supplementary  volume  to  the  one  upon 
mechanical  movements.  Unlike  the  first  volume,  which 
is  more  elementary  in  character,  this  volume  contains 
illustrations  and  descriptions  of  many  combinations 
of  motions  and  of  mechanical  devices  and  appliances 
found  in  different  lines  of  Machinery.  Each  device 
being  shown  t>v  a  line  drawing  with  a  description 
showing  its  working  parts  and  the  method  of  operation. 
396  pages,  1,000  specially  made  illustrations.  Price, 
$3.00. 

HISCOX.    Modern  Steam  Engineering  in  Theory  and 

Practice. 

This  book  has  been  specially  prepared  for  the  use 
of  the  modern  steam  engineer,  the  technical  students, 
and  all  who  desire  the  latest  snd  most  reliable  infor- 
mation on  steam  and  steam  boilers,  the  machinery  of 
power,  the  steam  turbine,  electric  power  and  lighting 
plants,  etc.  450  pages,  400  detailed  engravings.  $3.00. 

HOBART.    Brazing  and  Soldering. 

A  complete  course  of  instruction  in  a!l  kinds  of 
hard  and  soft  soldering.  Shows  just  what  tools  to  use, 
how  to  make  them  and  how  to  use  them.  Price,  25 
cents. 

HORNER.   Modern  Milling  Machines:  Their  Design, 

Construction  and  Operation. 

This  work  of  304  pages  is  fully  illustrated  and  de- 
scribes and  illustrates  the  Milling  Machine  in  every 
detail.  $4.00. 


GOOD,  PRACTICAL  BOOKS 


HORNER,    Practical  Metal  Turning. 

A  work  covering  the  modern  practice  of  machining 
metal  parts  in  the  lathe.  Fully  illustrated.  $3.50. 

HORNER.    Tools  for  Machinists  and  Wood  Work- 
ers, Including  Instruments  of  Measurement. 
A    practical    work    of    340    pages    fully    illustrated, 
giving  a  general  description  and  classification  of  tools 
for  machinists  and  woodworkers.    $3.50. 

Horse  Power  Chart. 

Shows  the  horse  power  of  any  stationary  engine 
without  calculation.  No  matter  what  the  cylinder 
diameter  or  stroke;  the  steam  pressure  or  cut-off;  the 
revolutions,  or  whether  condensing  or  non-condensing, 
it's  all  there.  Easy  to  use,  accurate  and  saves  time 
and  calculations.  Especially  useful  to  engineers  and 
designers.  50  cents. 

Inventor's  Manual;  How  to  Make  a  Patent  Pay. 

This  is  a  book  designed  as  a  guide  to  inventors  in 
perfecting  their  inventions,  taking  out  their  patents 
and  disposing  of  them.  119  pages.  Cloth,  $1.00. 

KLEINHANS.    Boiler  Construction. 

The  only  book  showing  how  locomotive  boilers  are 
built  in  modern  shops.  Shows  all  types  of  boilers 
used;  gives  details  of  construction;  practical  facts, 
such  as  life  of  riveting  punches  and  dies,  work  done 
per  day,  allowance  for  bending  and  flanging  sheets 
and  other  data  that  means  dollars  to  any  railroad 
man.  421  pages.  334  illustrations.  Six  folding  plates. 
$3.00. 

KRATTSS.   Linear  Perspective  Self-Taught. 

The  underlying  principle  by  which  objects  may  be 
correctly  represented  in  perspective  is  clearly  set  forth 
in  this  book;  everything  relating  to  the  subject  is 
shown  in  suitable  diagrams,  accompanied  by  full  ex- 
planations in  the  text.  Price.  $2.50. 

LE  VAN.    Safety  Valves;  Their  History,  Invention, 

and  Calculation. 

Illustrated  by  69  engravings.    151  pages.    $1.50. 
MARKHAM.    American  Steel  Worker. 

The  standard  work  on  hardening,  tempering  and 
annealing  steel  of  all  kinds.  A  practical  book  for  the 
machinist,  tool  maker  or  superintendent.  Shows  just 
how  to  secure  best  results  in  any  case  that  comes 
along.  How  to  make  and  use  furnaces  and  case 
harden;  how  to  handle  high-speed  steel  and  how  to 
temper  for  all  classes  of  work.  $2.50. 


GOOD,  PRACTICAL  BOOKS 


JIATHOT.    Modern  Gas  Engines  and  Producer  Gas 

Plants. 

A  practical  treatise  of  320  pages,  fully  illustrated 
by  175  detailed  illustrations,  setting  forth  the  princi- 
ples of  gas  engines  and  producer  design,  the  selection 
and  installation  of  an  engine,  conditions  of  perfect 
operation,  producer-gas  engines  and  their  possibilities, 
the  care  of  gas  engines  and  producer-gas  plants,  with 
a  chapter  on  volatile  hydrocarbon  and  oil  engines. 
$2.50. 

MEINHARDT.    Practical  Lettering  and  Spacing. 

Shows  a  rapid  and  accurate  method  of  becoming  a 
good  letterer  with  a  little  practice.  60  cents. 

PARSELL  &  WEED.    Gas  Engine  Construction. 

A  practical  treatise  describing  the  theory  and  prin- 
ciples of  the  action  of  gas  engines  of  various  types, 
and  the  design  and  construction  of  a  half-horse-power 
gas  engine,  with  illustrations  of  the  work  in  actual 
progress,  together  with  dimensioned  working  drawings 
giving  clearly  the  sizes  of  the  various  details.  300 
pages.  $2.50. 

PERRIGO.    Change  Gear  Devices. 

A  book  for  every  designer,  draftsman  and  mechanic 
who  is  interested  in  feed  changes  for  any  kind  of 
machines.  This  shows  what  has  been  done  and  how. 
Gives  plans,  patents  and  all  information  that  you 
need.  Saves  hunting  through  patent  records  and  rein- 
venting old  ideas.  A  standard  work  of  reference. 
$1.00. 

PEERIGO.    Modern  American  Lathe  Practice. 

A  new  book  describing  and  illustrating  the  very  lat- 
est practice  in  lathe  and  boring  mill  operations,  as 
well  as  the  construction  of  and  latest  developments 
in  the  manufacture  of  these  important  classes  of 
machine  tools.  300  pages,  fully  illustrated.  $2.50. 

PERRIGO.     Modern    Machine    Shop    Construction, 
Equipment  and  Management. 

The  only  work  published  that  describes  the  Modern 
Machine  Shop  or  Manufacturing  Plant  from  the  time 
the  grass  is  growing  on  the  site  intended  for  it  until 
the  finished  product  is  shipped.  Just  the  book  needed 
by  those  contemplating  the  erection  of  modern  shop 
buildings,  the  rebuilding  and  reorganization  of  old 
ones,  or  the  introduction  of  Modern  Shop  Methods, 
Time  and  Cost  Systems.  It  is  a  book  written  and  illus- 
trated by  a  practical  shop  man  for  practical  shop  men 
who  are  too  busy  to  read  theories  and  want  facts.  It 


GOOD,  PRACTICAL  BOOKS 


is  the  most  complete  all-around  book  of  its  kind  ever 
published.  400  large  quarto  pages,  225  original  and 
specially-made  illustrations.  $5.00. 

PRATT.    Wiring  a  House. 

Shows  every  step  in  the  wiring  of  a  modern  house 
and  explains  everything  so  as  to  be  readily  under- 
stood. Directions  apply  equally  to  a  shop.  Price,  25 
cents. 

REAGAN,  JR.    Electrical  Engineers'  and  Students' 
Chart  and  Hand-Book  of  the  Brush  Arc  Light 
System. 
Bound  in  cloth,  with  celluloid  chart  in  pocket.    50 

cents. 

RICHARDS  AND  COLYIN.    Practical  Perspective. 

Shows  just  how  to  make  all  kinds  of  mechanical 
drawings  in  the  only  practical  perspective  isometric. 
Makes  everything  plain  so  that  any  mechanic  can  un- 
derstand a  sketch  or  drawing  in  this  way.  Saves  time 
in  the  drawing  room  and  mistakes  in  the  shops.  Con- 
tains practical  examples  of  various  classes  of  work. 
60  cents. 

ROUILLION.    Drafting  of  Cams. 

The  laying  out  of  cams  is  a  serious  problem  unless 
you  know  how  to  go  at  it  right.  This  puts  you  on 
the  right  road  for  practically  any  kind  of  cam  you 
are  likely  to  run  up  against.  Price,  25  cents. 

ROUILLION.   Economics  of  Manual  Training. 

The  only  book  that  gives  just  the  information  need- 
ed by  all  interested  in  manual  training,  regarding 
buildings,  equipment  and  supplies.  Shows  exactly 
what  is  needed  for  all  grades  of  the  work  from  the 
Kindergarten  to  the  High  and  Normal  School.  Gives 
itemized  lists  of  everything  needed  and  tells  just 
what  it  ought  to  cost.  Also  shows  where  to  buy  sup- 
plies. $2.00. 

SATTNIER.    Watchmaker's  Hand-Book. 

Just  issued,  7th  edition.  Contains  498  pages  and  is 
a  workshop  companion  for  those  engaged  in  watch- 
making and  allied  mechanical  arts.  250  engravings  and 
14  plates.  $3.00. 

SLOANE.     Electricity  Simplified. 

The  object  of  "Electricity  Simplified"  is  to  make 
the  subject  as  plain  as  jpossible  and  to  show  what  the 
modern  conception  of  electricity  is.  158  pages.  $1.00. 


GOOD,  PRACTICAL  BOOKS 


SLOANE.   How  to  Become  a  Successful  Electrician. 

It  is  the  ambition  of  thousands  of  young  and  old 
to  become  electrical  engineers.  Not  every  one  is  pre- 
pared to  spend  several  thousand  dollars  upon  a  col- 
lege course,  even  if  the  three  or  four  years  requisite 
are  at  their  disposal.  It  is  possible  to  become  an  elec- 
trical engineer  without  this  sacrifice,  and  this  work  is 
designed  to  tell  "How  to  Become  a  Successful  Elec- 
trician" without  the  outlay  usually  spent  in  acquiring 
the  profession.  189  pages.  $1.00. 

SLOANE.    Arithmetic  of  Electricity. 

A  practical  treatise  on  electrical  calculations  of  all 
kinds  reduced  to  a  series  of  rules,  all  of  the  simplest 
forms,  and  involving  only  ordinary  arithmetic;  each 
rule  illustrated  by  one  or  more  practical  problems, 
with  detailed  solution  of  each  one.  138  pages.  $1.00. 

SLOANE.    Electrician's  Handy  Book. 

An  up-to-date  work  covering  the  subject  of  practical 
electricity  in  all  its  branches,  being  intended  for  the 
every-day  working  electrician.  The  latest  and  best 
authority  on  all  branches  of  applied  electricity.  Pocket- 
book  size.  Handsomely  bound  in  leather,  with  title 
and  edges  in  gold.  800  pages.  500  illustrations.  Price, 
$3.50. 

SLOANE.    Electric  Toy  Making,  Dynamo  Building, 
and  Electric  Motor  Construction. 

This  work  treats  of  the  making  at  home  of  electrical 
toys,  electrical  apparatus,  motors,  dynamos,  and  in- 
struments in  general,  and  is  designed  to  bring  within 
the  reach  of  young  and  old  the  manufacture  of  genu- 
ine and  useful  electrical  appliances.  140  pages.  $1.00. 

SLOANE.    Rubber  Hand  Stamps  and  the  Manipula- 
tion of  India  Rubber. 

A  practical  treatise  on  the  manufacture  of  all  kinds 
of  lubber  articles.  146  pages.  $1.00. 

SLOANE.     Liquid    Air    and    the    Liquefaction    of 
Gases. 

Containing1  the  full  theory  of  the  subject  and  giving 
the  entire  history  of  liquefaction  of  gases  from  the 
earliest  times  to  the  present.  365  pages,  with  many 
illustrations.  $2.00. 

SLOANE.    Standard  Electrical  Dictionary. 

A  practical  handbook  of  reference,  containing  defi- 
nitions of  about  5,000  distinct  words,  terms,  and 
phrases.  682  pages.  393  illustrations.  $3.00. 


GOOD,  PRACTICAL  BOOKS 


STARBUCK,    Modern  Plumbing  Illustrated. 

A  comprehensive  and  up-to-date  work  illustrating 
and  describing  the  Drainage  and  Ventilation  of  dwell- 
ings, apartments,  and  public  buildings,  etc.  Adopted 
by  the  United  States  Government  in  its  sanitary  work 
in  Cuba,  Porto  Rico,  and  the  Philippines,  and  by  the 
principal  boards  of  health  of  the  United  States  and 
Canada.  The  standard  book  for  master  plumbers, 
architects,  builders,  plumbing  inspectors,  boards  of 
health,  boards  of  plumbing  examiners,  and  for  the 
property  owner,  as  well  as  for  the  workman  and  his 
apprentice.  300  pages.  55  full-page  illustrations. 
$4.00. 

Tonnage  Chart, 

Built  on  the  same  lines  as  the  Tractive  Power 
Chart;  it  shows  the  tonnage  any  tractive  power  will 
haul  under  varying  conditions  of  road.  No  calcula- 
tions are  required.  Knowing  the  drawbar  pull  and 
grades  and  curves  you  find  tonnage  that  can  be 
hauled.  50  cents. 

Tractive  Power  Chart. 

A  chart  whereby  you  can  find  the  tractive  power 
or  drawbar  pull  of  any  locomotive,  without  making  a 
figure.  Shows  what  cylinders  are  equal,  how  driving 
wheels  and  steam  pressure  affect  the  power.  What 
sized  engine  you  need  to  exert  a  given  drawbar  pull 
or  anything  you  desire  in  this  line.  Printed  on  tough 
jute  paper  to  stand  rolling  or  folding.  50  cents. 

USHER.    The  Modern  Machinist. 

A  practical  treatise  embracing  the  most  approved 
methods  of  modern  machine-shop  practice,  and  the  ap- 

Slications    of    recent    improved    appliances,    tools,    and 
evices    for    facilitating,    duplicating,    and    expediting 
the    construction    of    machines    and    tlip.ir   parts.     257 
engravings.    322  pages.    $2.50 

VAN  DERVOORT,  Modern  Machine  Shop  Tools; 
Their  Construction,  Operation,  and  Manipula- 
tion. 

An  entirely  new  and  fully  illustrated  work  of  555 
pages  and  673  illustrations,  describing  in  every  de- 
tail the  construction,  operation,  and  manipulation  of 
both  Hand  and  Machine  Tools.  Includes  chapters  on 
filing,  fitting,  and  scraping  surfaces;  on  drills,  ream- 
ers, taps,  and  dies;  the  lathe  and  its  tools;  planers, 
shapers,  and  their  tools;  milling  machines  and  cutters; 
gear  cutters  and  gear  cutting;  drilling  machines  and 
drill  work;  grinding  machines  and  their  work;  hard- 
ening and  tempering;  gearing,  belting,  and  transmis- 
sion machinery;  useful  data  and  tables,  $4,00, 


GOOD,  PRACTICAL  BOOKS 


WALLIS-TAYLOR.    Pocket  Book  of  Refrigeration 
and  Ice-Making. 

This  is  one  of  the  latest  and  most  comprehensive 
reference  books  published  on  the  subject  of  refriger- 
ation and  cold  storage.  $1.50. 

WOOD.    Walschaert  Locomotive  Valve  Gear. 

The  only  work  issued  treating  of  this  subject  of 
valve  motion.  150  pages.  $1.50. 

WOODWORTH.    American  Tool  Making  and  Inter- 
changeable Manufacturing. 

A  practical  treatise  of  560  pages,  containing  600 
illustrations  on  the  designing,  constructing,  use,  and 
installation  of  tools,  jigs,  fixtures,  devices,  special 
appliances,  sheet-metal  working  processes,  automatic 
mechanisms,  and  labor-saving  contrivances;  together 
with  their  use  in  the  lathe,  milling  machine,  turret 
lathe,  screw  machine,  boring  mill,  power  press,  drill, 
subpress,  drop  hammer,  etc.,  for  the  working  of  met- 
als, the  production  of  interchangeable  machine  parts, 
and  the  manufacture  of  repetition  articles  of  metal. 
$4.00. 

WOODWORTH.    Dies,  Their  Construction  and  Use 
for  the  Modern  Working  of  Sheet  Metals. 

A  new  book  by  a  practical  man,  for  those  who  wish 
to  know  the  latest  practice  in  the  working  of  sheet 
metals.  It  shows  how  dies  are  designed,  made  and 
used,  and  those  who  are  engaged  in  this  line  of  woris 
can  secure  many  valuable  suggestions.  Thoroughly 
modern.  384  pages.  505  illustrations.  $3.00. 

WOODWORTH.    Hardening,  Tempering,  Annealing, 
and  Forging  of  Steel. 

A  new  book  containing  special  directions  for  the 
successful  hardening  and  tempering  of  all  steel  tools. 
Milling  cutters,  taps,  thread  dies,  reamers,  both  solid 
and  shell,  hollow  mills,  punches  and  dies,  and  all  kinds 
of  sheet-metal  working  tools,  shear  blades,  saws,  fine 
cutlery  and  metal-cutting  tools  of  all  descriptions,  as 
well  as  for  all  implements  of  steel,  both  large  and 
small,  the  simplest  and  most  satisfactory  hardening 
and  tempering  processes  are  presented.  The  uses  to 
which  the  leading  brands  of  steel  may  be  adapted  are 
concisely  presented,  and  their  treatment  for  working 
under  different  conditions  explained,  as  are  also  the 
special  methods  for  the  hardening  and  tempering  of 
special  brands,  320  pages.  250  illustrations.  $2.50. 


JUST     PUBLISHED 


HYDRAULIC  ENGINEERING 

By  Gardner  D.  Hiscox,  a  practical  work  on 
hydraulics  and  hydrostatics  in  principle  and  prac- 
tice, including  chapters  on : 

The  measurement  of  water  flow  for  power  arid 
other  uses;  siphons,  their  use  and  capacity;  hydrau- 
lic rams;  dams  and  barrages;  reservoirs  and  their 
construction;  city, town  and  domestic  water  sup- 
ply; wells  and  subterranean  water  flow;  artesian 
wells  and  the  principles  of  their  flow;  geological 
conditions ;  irrigation  water  supply,  resources  and 
distribution  in  arid  districts;  the  great  projects  for 
irrigation. 

Water  power  in  theory  and  practice,  water  wheels 
and  turbines;  pumps  and  pumping  devices,  cen- 
trifugal, rotary  and  reciprocating;  air-lift  and  air- 
pressure  devices  for  water  supply — hydraulic  power 
and  high-pressure  transmission ;  hydraulic  mining; 
marine  hydraulics,  buoyancy  displacement,  tonnage, 
resistance  of  vessels  and  skin  friction.  Relative 
velocity  of  waves  and  boats ;  tidal  and  wave  pow- 
er, with  over  300  illustrations  and  36  tables  of 
hydraulic  effect.  A  most  valuable  work  for  study 
and  reference.  About  400  octavo  pages.  Price, 
$4.00. 

THE  TELEPHONE  HAND-BOOK 

By  H.  C.  Gushing,  Jr.,  E.E.,  and  W.  H.  Radcliffe, 
E.E.  A  practical  reference  book  and  guide  for  tele- 
phone wiremen  and  contractors.  Every  phase  of 
telephone  wiring  and  installation  commonly  used 
to-day  is  treated  in  a  practical,  graphic  and  concise 
manner.  Fully  illustrated  by  half-tones  and  line- 
cut  diagrams,  showing  the  latest  methods  of  install- 
ing and  maintaining  telephone  systems  from  the 
simple  two-instrument  line,  intercommunication 
systems  for  factories,  party  lines,  etc.  175  pages, 
100  illustrations,  cloth,  pocket  size.  $1.00. 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 


AN     INITIAL    FINE     OF     25     CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  5O  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.OO  ON  THE  SEVENTH  DAY 
OVERDUE. 


6  1932 


LD  21- 


7B  0958' 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 


